If we add N and the sum of its digits, then we get 313.
What are the possible values of N?
How to solve the puzzle?
Let's start with the largest two-digit number i.e. 99. Let’s add 99 and the sum of its digits:
99 + 9 + 9 = 117
Therefore, N must be a 3-digit number. Also, N cannot exceed 313.
Next, the maximum sum of digits of a 3-digit number is 27.
Therefore, N must at least be 313-27 = 286.
286 < N < 313
We can say that N is either of the form 28p or 29p or 30p or 31p (where p is the unit’s digit).
For the form 28p:
28p + (2 + 8 + p) = 313
280 + 10 + 2p = 313
This is not possible since the LHS is clearly an even number, while 313 is odd.
For the form 29p:
29p + (2 + 9 + p) = 313
290 + 11 + 2p = 313
290 + 2p = 313-11 = 302
2p = 302 – 290 = 12
p = 12/2 = 6
Therefore, the first possible value of N is 296.
For the form 30p:
30p + (3 + 0 + p) = 313
300 + 3 + 2p = 313
2p = 313 – 300 – 3 = 10
p = 10/2 = 5
Thus, the second possible value of N is 305.
For the form 31p:
31p + (3 + 1 + p) = 313
310 + 4 + 2p = 313
314 + 2p = 313
This is not possible since the LHS is clearly greater than 313 (314 is greater than 313 and N has to be a positive integer).
So, we have two possible values of N, 296, and 305.
Hope you liked the challenge! At Cuemath, we regularly present challenging and stimulating math puzzles to our students so that their logical thinking and problem-solving abilities are enhanced.
**This puzzle appeared in the book – The Ultimate Mathematical Challenge Over 365 puzzles to test your wits and excite your mind, written by Marcus du Sautoy, a Professor of Mathematics and Simonyi Professor for the Public Understanding of Science at the University of Oxford.
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