SOLUTION: Hi everyone, please can you help me with a maths problem: I am a odd 3 digit number, that is equal to another number multiplied by itself. the sum of my digits is 16. what number

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Question 1024875: Hi everyone, please can you help me with a maths problem: I am a odd 3 digit number, that is equal to another number multiplied by itself. the sum of my digits is 16. what number am i.
Thanks!!!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The number that is multiplied by itself can't be a
1-digit number. It can't be a 3-digit number, so
it must be a 2-digit number.
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Also, the units digit must be odd since
( odd )x( odd) = odd in order to make +c+ odd
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It can't be as large as +33+, since +33%5E2+=+1089+
I could be as large as +31+, since +31%5E2+=+961+
It could be as small as +11+, since +11%5E2+=+121+
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So, the possible numbers and their squares are:
11 - 121
13 - 169
15 - 225
17 - 289
19 - 361
21 - 441
23 - 529
25 - 625
27 - 729
29 - 841
31 - 961
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The sum of the digits of +13%5E2+=+169+ is +16+
The sum of the digits of +23%5E2+=+529+ is +16+
The sum of the digits of +31%5E2+=+961+ is +16+
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I think the number can be
169, 529, or 961