Mathematical Challenges

 

 

Number Properties

 

  1.  Find the smallest 8-digit number, no two digits of which are the same, that is divisible by 9.

Solution 1 by Foong Ruyi (Sec 1, 2003), Bukit Panjang Government High Sch

Ruyi provided the idea behind the following solution:

The smallest 8-digit number, in which no two digits are the same, is the number 10234567 whose digital sum is 1+0+2+3+4+5+6+7 = 28.

To be divisible by 9, the digital sum must be a multiple of 9. Since the next multiple of 9 greater than 28 is 36, some digits of the number 10234567 have to be replaced until the digital sum is 36. In other words, the digital sum has to be increased by 8.

Starting from the last digit, we replace 7 by 9, increasing the digital sum by 2. Then we replace the second last digit 6 by 8, again increasing the digital sum by 2, and so on until the digital sum is increased by 8. The reason why we choose to start from the last digit is to keep the remaining digits as small as possible. Note that each time we can only increase the digital sum by 2, to avoid repetition of digits.

The required smallest 8-digit number is therefore 10236789.

Solution 2 by Chan Hui Mui (Sec 1, 2004), Hillgrove Sec Sch

Hui Mui provided the idea behind the following solution:

The sum of the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 is 45. So, any 10-digit number formed by the ten digits will be divisible by 9. To form an 8-digit number that is divisible by 9, we need to remove two of the digits such that the sum of the remaining eight digits is a multiple of 9. This can be done by removing any two digits which add up to 9, so that the sum of the remaining 8 digits is 45 – 9 = 36. In order that the 8-digit number is as small as possible, we remove the digits 4 and 5. The smallest number that can be formed by the remaining digits is 10236789.

We can show that this is the smallest answer by considering the other possibilities. For example, if we remove the digits 3 and 6, the smallest number that can be formed from the remaining digits is 10245789. Similarly, after removing 2 and 7 the smallest number that can be formed is 10345689. Finally, if we remove 1 and 8, the smallest number that can be formed is 20345679. These are all larger than 10236789.

 

 

  2.  Find the smallest positive integer that contains each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once, and is divisible by each of the numbers 2, 3, 4, 5, 6, 7, 8, 9.

Solution by Lin Chuan Wen & Tian Shi Xin Sophil (Sec 1, 2004), Bukit Panjang Government High Sch

Chuan Wen & Shi Xin provided the idea behind the following solution:

Consider numbers of the form 1234******, where the remaining digits 4, 5, 6, 7, 8, 9 and 0 occupy the positions of the asterisks. The last digit must be 0, as the number must be divisible by 2 and 5. So the number must be of the form 1234*****0.

We do not have to consider divisibility by 3 and 9, since the digital sum (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 = 45) is divisible by 3 and 9. Also, since the number is divisible by 2 and 3, it is divisible by 6. So, we only have to consider divisibility by 4, 7 and 8.

First, consider divisibility by 4, we place the digit 8 in the tens position so that the number is as small as possible. Obviously, we can’t put the digit 9 there because 90 is not divisible by 4. So, the number must be of the form 1234****80.

Next, consider divisibility by 8. The digit 6 should occupy the hundreds position as it is the largest available at this point, since 980 and 780 are not divisible by 8. The number is now reduced to the form 1234***680. To make the number as small as possible, we place the digit 9 at the thousands position. The number is further reduced to the form 1234**9680, with the remaining digits 5 and 7 occupying the positions of the asterisks.

Of the two numbers 1234579680 and 1234759680 only the latter is divisible by 7. Hence, the smallest integer that contains each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once and is divisible by each of the numbers 2, 3, 4, 5, 6, 7, 8, 9 is 1234759680.

 

 Remarks:

(1) This number is also divisible by 11 and 12.

(2) There are 1056 integers that contain each of the digits 0 through 9 exactly once, and are divisible by each of the numbers 2 through 12. If you are interested to know what these 1056 numbers are, you may run this small program written in Liberty Basic: highdiv.bas. However, you need to install the Liberty Basic software. Click here to download the software.

 

 

  3.  Arrange the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form a nine-digit number such that the number formed by the first two digits is divisible by 2, the number formed by the first three digits is divisible by 3, the number formed by the first four digits is divisible by 4, the number formed by the first five digits is divisible by 5, and so on, until the entire nine-digit number is divisible by 9. [Martin Gardner]

Solution by Alexander George Wood (Sec 2, 2006), Tanjong Katong Sec Sch

Point 1     For the number formed by the first 2 digits, first 4 digits, first 6 digits, and first 8 digits to be divisible by 2, 4, 6, and 8 respectively, the 2nd, 4th, 6tth and 8th digits all have to be even. Therefore all the four even digits (2, 4, 6, 8) are used in all the four even positions. This results in the even positions having even digits and the odd positions having odd digits.

Point 2     The number formed by the first five digits is divisible by 5. Therefore the 5th digit should be 5, as there is no 0.

Point 3     The whole 9-digit number has to be divisible by 9. This means the sum of the digits has to be a multiple of 9. Since 1+2+3+4+5+6+7+8+9 = 45 is divisible by 9, this means that no matter what order we arrange the digits the number formed will always be divisible by nine. So, we are able to ignore the divisibility test for nine.

Point 4     The number formed by the first 3 digits is divisible by 3, so the sum of the digits has to be divisible by 3. Also, the 1st and 3rd digits are odd, and the 2nd digit must be even (proven in point 1). This leads to the following combinations of first three digits:

                123, 129, 147, 183, 189   (cannot be 141 as the digit 1 is used twice; also, cannot be 165 because the number 5 has been used [point 2])

                321, 327, 369, 381, 387   (cannot be 345, 363)

                723, 729, 741, 783, 789   (cannot be 747, 765)

                921, 927, 963, 981, 987   (cannot be 945, 969)

Point 5     The number formed by the first 4 digits is divisible by 4, so the 3rd and 4th digits taken as a 2-digit number is divisible by 4. Since the 3rd digit is odd and the 4th digit is even (proven in point 1), we shall list all the 2-digit numbers whose first digit is odd and second digit is even and are divisible by 4. These are:

                12, 16, 32, 36, 72, 76, 92, and 96.

                (The numbers 52 and 56 are not possible as the digit 5 has to be in the 5th position [point 2].)

Point 6     Points 4 and 5 together will form possible combinations of the 1st to the 4th digits: (Combinations with repeating digits have been removed.)

                1236, 1296, 1472, 1476, 1832, 1836, 1892, 1896,

                3216, 3276, 3692, 3812, 3816, 3872, 3876,

                7236, 7296, 7412, 7416, 7832, 7836, 7892, 7896,

                9216, 9276, 9632, 9816, 9812, 9872, 9876.

Point 7     The number formed by the first 6 digits has to be divisible by 6, so they must be divisible by 2 and 3. The sum of the first 3 digits is already divisible by 3, so we only need to take care of the 4th, 5th and 6th digits. The 4th and 6th digits have to be even (see point 1). The 5th digit has to be 5 (see point 2). The possible combinations are:

                258, 456, 654, 852 (252 and 858 are not possible)

Point 8     Putting Point 6 and Point 7 together makes these possible combinations of the 1st to 6th digits (combinations with repeating digits have been removed):

                123654, 129654, 147258, 183654, 189654,

                321654, 327654, 369258, 381654, 387654,

                723654, 729654, 741258, 783654, 789654,

                921654, 927654, 963258, 981654, 987654.

Point 9     The number formed by the first 8 digits is divisible by 8, which means that the 6th, 7th and 8th digit taken together as a 3-digit number must be divisible by 8. The 6th and 8th digits are even while the 7th one is odd (point 1). Also 5 cannot be used as it has already been used (point 2). So the possible combination for the 6th, 7th and 8th digits are:  216, 296, 416, 432, 472, 496, 632, 672, 816, 832, 872, and 896.

Point 10   Combining points 8 and 9, the possible combinations of 1st to 8th digit are:

                14725896, 18365472, 18965432, 18965472, 38165472,

                74125896, 78965432, 98165432, 98165472, 98765432.

Point 11   Now we add in the last digit of the 9-digit number.  Referring to point 3, all we have to do is pop in the last digit available since we do not need to check for divisibility of 9.

                147258963, 183654729, 189654327, 189654723, 381654729,

                741258963, 789654321, 981654327, 981654723, 987654321.

Point 12   We have already taken care of the divisibility of 2, 3, 4, 5, 6, 8 and 9. In this final point, we shall check the divisibility of 7. The number which passes this test is the final answer:

                Taking the first 7 digits of each number listed in point 11 and dividing them by 7, the only number which passes the divisibility test of 7 is 381654729. It is therefore the unique answer to the question.

 

Remarks:

This is a very clearly presented solution, with all the steps logically explained. Can you improve on the efficiency of the solution?

 

 

  4.  The product of two positive integers is 588000. Find the largest possible value for the highest common factor of the two integers.

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

Yishun provided the idea behind the following solution:

First, factorise the number into prime factors to get 588000 = 2⁵ ´ 3 ´ 5³ ´ 7². Then recombine the prime factors, expressing the number as the product of two integers, say x and y. To obtain the largest possible value for the highest common factor (HCF) of x and y, we distribute the prime factors with powers of 2 and above as evenly as possible between x and y. Since there are five 2’s, we distribute two 2’s each to x and y and leave out the remaining 2. Similarly, we distribute one 5 and one 7 each to x and y. This is summarized in the table below:

The remaining prime factors in the last column can be distributed to x and y in any way as it will not change the value of the HCF. For example, we can have x = 2 ´ 2 ´ 5 ´ 7, y = 2 ´ 2 ´ 5 ´ 7 ´ 2 ´ 3 ´ 5 or x = 2 ´ 2 ´ 5 ´ 7 ´ 5, y = 2 ´ 2 ´ 5 ´ 7 ´ 2 ´ 3, etc.

Therefore, the largest possible value for the HCF is 2 ´ 2 ´ 5 ´ 7 = 140.

 

 

  5.  Fill in the boxes with the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 to make the equation true. Each digit must be used exactly once. Show that the solution is unique.

5568 =  ´  =  ´ 

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

Yishun provided the idea behind the following solution:

First, factorise the number into prime factors to get 5568 = 2⁶ ´ 3 ´ 29. Then recombine the prime factors, expressing the number as the product of two integers which are two-digit or three-digit. There are eight possible pairs:

12 ´ 464, 16 ´ 348, 24 ´ 232, 29 ´ 192, 32 ´ 174, 48 ´ 116, 58 ´ 96, 64 ´ 87

Eliminating all pairs with repeated digits, we are left with the following 4 possible pairs:

16 ´ 348, 32 ´ 174, 58 ´ 96, 64 ´ 87

Since each of the three-digit integers contains the digit 4, we eliminate the pair 64 ´ 87 as well. So, one of the required pair is 58 ´ 96. The remaining pair must be 32 ´ 174, in order to have no repetition of digits. We conclude that the solution found is unique.

5568 = 58 ´ 96 = 32 ´ 174

 

 

Digit Problems

 

  6.  A two-digit number is squared. When this two-digit number is reversed and squared, the difference between the squares is also a square. What is the two-digit number?

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

Let a, b be the tens digit and the ones digit respectively. We may assume a > b.

To be a perfect square, a² – b² must be 11 times of a square number. Let a² – b² = 11k².

If k = 1,

  

then a – b = 1 and a + b = 11.

We have a = 6, b = 5.

If k = 2,

    = 1 ´ 44 = 2 ´ 22 = 4 ´ 11,

then a – b = 1, a + b = 44 or a – b = 2, a + b = 22 or a – b = 4, a + b = 11. None of these pairs of equations leads to a valid solution.

If k = 3, then a² – b² = 99. There is no solution in this case as the largest value of a² – b² is 80 (when a = 9 and b = 1). Similarly, there is no solution for larger values of k.

Therefore, the only possible answer is 65.

 

 

  7.  How many four-digit numbers of the form aabb are perfect squares?

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

To be a perfect square, 100a + b must be 11 times a square number, i.e., 100a + b = 11k² for some integer k. Also, 100a + b is of the form a0b, that is, a three-digit number with a 0 at the middle.

All the 3-digit multiples of 11 with a 0 at the middle are: 209, 308, 407, 506, 605, 704, 803, 902. To check which one is 11 times a square number we divide each number in turn by 11.

The results are, respectively: 19, 28, 37, 46, 64 (a square number), 73, 82, 91.

Therefore 100a + b = 704, from which a = 7 and b = 4. We conclude that 7744 is the unique answer.

 

Remarks:

We could also have examined numbers of the form 11k² and see which one gives a 3-digit number with a 0 at the middle. E.g., 11(4)² = 176, 11(5)² = 275, 11(6)² = 396, 11(7)² = 539, 11(8)² = 704 (this is the one we are looking for), 11(9)² = 891.

 

 

  8.  The integer N is 13 times the sum of its digits. Find all the possible values of N.

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

Yishun provided the idea behind the following solution:

We can show that N cannot be a 4-digit number. If N is the number abcd, then

13 (a + b + c + d) £ 13 ´ 9 ´ 4 = 468 < 1000.

Similarly, larger numbers are not possible.

We can also show that N cannot be a 2-digit number. If N is the number ab, then

This is not possible since a > 0 and b ³ 0.

So the answers can only be 3-digit numbers. Let N be abc, then

If a = 1, the equation becomes 29 = b + 4c. We have the following sets of solutions:

If a = 2, the equation becomes 58 = b + 4c. This has no solution as the largest value of b + 4c is 45 (when b = 9 and c = 9).

Similarly, there is no solution for a > 2.

Therefore, the possible values of N are 117, 156 and 195.

 

 

Mathematical Titbits

 

  9.  The integer n is such that n > 2 and (n!)²(n² – 2)! =  (n²)![(n – 2)!]², find the value of n.

Solution by Adelene Tan Xiu Wen (Sec 2, 2004), Bukit Panjang Government High Sch

 

 

Indices

 

10.  Which is larger, 16¹⁸ or 18¹⁶?

Solution 1 by Xiong Yishun (Sec 2, 2005), Bukit Panjang Government High Sch

         16¹⁸ = (2⁴)¹⁸ = 2⁷² = (2⁹)⁸ = 512⁸

         18¹⁶ = (18²)⁸ = 324⁸

Therefore, 16¹⁸ > 18¹⁶.

Solution 2 by Xiong Yishun (Sec 2, 2005), Bukit Panjang Government High Sch

For any two positive numbers A and B, if  then A > B.

        

Therefore, 16¹⁸ > 18¹⁶.

 

 

Geometry

 

11.  In DABC, D, E, F are points on the sides BC, CA and AB respectively such that AF = 3FB, BD = 4DC and CE = 5EA. If the area of DABC = 120, find the area of DDEF.

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

 

Remarks:

In the solution below the notation [XYZ] is used to represent the area of triangle XYZ. This will make the working neater and easier to follow.

 

 

12.  A triangle is cut into 3 smaller triangles and a quadrilateral. The areas of the smaller triangles are 6, 15 and 30 square units as shown. What is the area of the quadrilateral?

Solution by Xiong Yishun (Sec 1, 2004), Bukit Panjang Government High Sch

 

Remarks:

In the solution below the notation [XYZ] is used to represent the area of triangle XYZ. This will make the working neater and easier to follow.

Label the vertices A, B, C, D, E and F as shown in the diagram. DECF and DBCF have the same height, so the ratio of their bases is the same as the ratio of the areas, i.e., . Also, DEDF and DBDF have the same height. Therefore, [EDF] =  [BDF] = 3.
Next, observe that DACD and DBCD have the same height. Furthermore, DAED and DBED have the same height and share the common bases AD and BD with the previous pair. Let x be the area of DAED. We have, Solving the equation, we get x = 6.

Thus, the area of quadrilateral ADFE = 3 + 6 = 9.

 

 

Trigonometry

 

13.  Given that sin⁶ x + cos⁶ x =  and 0° < x < 90°, find the value of 100 sin x cos x.

Solution by Teo Meng How (Sec 4, 2005), Swiss Cottage Sec Sch

Also solved by Lin Jinjie (Sec 4, 2005), Fuhua Sec Sch

 

 

14.  In the diagram, triangle ABC has a right angle at B. The point P inside the triangle is such that PA = 10, PB = 6 and ÐAPB = ÐBPC = ÐCPA. Find PC.

Solution by Ian Teoh Yuan Hao (Sec 4, 2005), Swiss Cottage Sec Sch

From DAPB,

                       

From DABC,

              

From DAPC,

                       

From DBPC,

                       

Substituting (2) and (3) into (1),

              

 

 

 

Intermediate Algebra

 

15.  Find the smallest positive integer n such that .

Solution by Quek Jia Hao (Sec 4, 2005), Fuhua Sec Sch

Let x = 4n – 1, then

              

Squaring both sides,

              

Squaring both sides again,

              

Therefore, the smallest integral value of n is 626.

 

Remarks:

(1) You may also use the substitution x = 4n – 2. The working will be slightly easier.

(2) Can you find a simpler solution?

 

16.  Given that x =  – 1, find the value of x¹⁰ + 2x⁹ – 2x⁸ – 2x⁷ + x⁶ + 3x² + 6x + 1.

Solution by Sze Swai Ming (Sec 4, 2006), Fuhua Sec Sch

Given x =  – 1, then x + 1 = .

Squaring both sides,

              

Therefore,

              

 

17.  Find the sum of all the real roots of the equation x⁴ – 4x³ + 5x² – 4x + 1 = 0.

Solution by Teo Meng How (Sec 4, 2005), Swiss Cottage Sec Sch

              

Let y = , then y² = . Thus,  = y² – 2.

Substituting into (1),

              

When y = 1,       

Discriminant = (–1)² – 4(1)(1) = –3 < 0. The equation has no real roots.

When y = 3,       

Therefore, sum of real roots =  = 3.

Remarks:

If a and b are the roots of the quadratic equation ax² + bx + c = 0, then the sum of roots, a + b = –b/a. Using this formula, we can find the sum of roots of a quadratic equation without finding the actual roots. Thus, the sum of roots of the equation x² – 3x + 1 = 0 is –(–3)/1 = 3.

 

 

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