# Can you solve this number puzzle?

**Hello! Thanks again for visiting!**

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.