Example 1. Tara thought of a whole number and then multiplied it by either 5 or 6. Krishnan added 5 or 6 to Tara’s answer. Finally, Shan subtracted either 5 or 6 from Krishnan’s answer. The final result was 73. What number did Tara choose?
Example 2. You are given that n is a positive integer with the property that when we add n and the sum of its digits, we obtain the number 313. What are the possible values of n?
Of course, you can find the solution just below, but it is highly recommended that you first try to solve it on your own.
Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.
Solution 1. Suppose Tara’s original number is x.
When she multiplied it, she obtained either 5x or 6x.
When Krishnan added 5 or 6, his answer was one of 5x + 5, 5x + 6, 6x + 5 or 6x + 6.
Finally, when Shan subtracted 5 or 6, his answer was one of 5x − 1, 5x, 5x + 1, 6x – 1, 6x + 1, or 6x.
The final result was 73. Obviously, 5x – 1 = 73 or 5x = 73 or 5x + 1 = 73 or 6x – 1 = 73 or 6x = 73 are not possible.
We consider 6x + 1 = 73, so that x = 12 is the required number.
Solution 2. As the largest sum of a two-digit number and its digits is 99 + 9 + 9 = 117, n must be a three-digit number.
Also, n is at most 313. Since the sum of the digits of a three-digit number is at most 27, n is at least 313 − 27 = 286.
So n is either 28d or 29d or 30d or 31d, where d is a digit.
If n = 28d, the sum of n and its digits is 280 + 10 + 2d, which is even and so cannot be 313.
If n = 29d, the sum of n and its digits is 290 + 11 + 2d, which equals 313 when d = 6. Hence 296 is a possible value for n.
If n = 30d, the sum of n and its digits is 300 + 3 + 2d, which equals 313 when d = 5.
Hence 305 is a possible value for n.
If n = 31d, the sum of n and its digits is 310 + 4 + 2d, which is even and so cannot be 313.
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