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The sum of the squares of the digits of a two digit number is 65. How 30 Dec 2019, 10:24
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

Question Stats:

84% (02:24) correct 16% (02:13) wrong based on 94 sessions

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The sum of the squares of the digits of a two digit number is 65. How much should be added to this number to get another two digit number in which the digits are in the reverse order?

(A) 59
(B) 25
(C) 63 or 27
(D) 21
(E) 32­
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C
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The sum of the squares of the digits of a two digit number is 65. How 30 Dec 2019, 10:42
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Given this specific option choices, this question can be solved without any calculation.

    • For any two-digit number, the number and its reverse will always have a gap, which is a multiple of 9 [if we have AB, then reverse of it is BA. BA – AB = 10B + A – 10A – B = 9(B – A)]

From the given choices, only one option is a multiple of 9

Hence, the correct answer is option C

From observation, the numbers are 18 (18 + 63 = 81) or 47 (47 + 27 = 74)
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The sum of the squares of the digits of a two digit number is 65. How 30 Dec 2019, 11:29
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First let's take a look at the squares of figures:

1 = 1
2 = 4
3= 9
4 = 16
5 = 25
6 = 36
7 = 49
8 = 64

There are a 2 combinations that gives 65: 1 & 8, 4 & 6; hence the possibilities are:

18 to 81, or 46, 64; hence answer is (C).
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The sum of the squares of the digits of a two digit number is 65. How 30 Dec 2019, 12:47
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naju
The sum of the squares of the digits of a two digit number is 65. How much should be added to this number to get another two digit number in which the digits are in the reverse order?
(A) 59
(B) 25
(C) 63 or 27
(d) 21
(E) 32

Let us assume, two digit number to be of the format xy where x is tenth digit and y is unit digit.
x^2 + y^2 = 65
(x,y) = {(1,8),(4,7)}

If xy = 18 -> yx=81; 81-18 = 63
If xy = 47 -> yx =74; 74-47 = 27

IMO C
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The sum of the squares of the digits of a two digit number is 65. How 24 Feb 2020, 02:20
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naju
The sum of the squares of the digits of a two digit number is 65. How much should be added to this number to get another two digit number in which the digits are in the reverse order?
(A) 59
(B) 25
(C) 63 or 27
(d) 21
(E) 32

Method 1: x = BA (the larger) - AB (the smaller) --> x = 10B + A - 10A - B = 9(B-A) --> x must be a multiple of 9 --> (C)
Method 2: A^2 + B^2 = 65 and A<B --> B max = 8.
(1) If B = 8, A = 1. --> x = 81 - 18 = 63
(2) If B = 7, A = 4 --> x = 74 - 47 = 27
--> (C)
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The sum of the squares of the digits of a two digit number is 65. How 27 Feb 2024, 13:05
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