Can you solve this number puzzle?

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Let’s start with the problem statement:

N is a positive integer.

If we add N and the sum of its digits, then we get 313.

What are the possible values of N?

Solution:

There are two possible values of N that satisfy the given condition: 296 and 305

Let’s see how the answer is derived in a logical way:

The largest two digit number is 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.

Hence,

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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