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The sum of the squares of the digits of a two digit number is 65. How

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Intern
Joined: 20 Aug 2019
Posts: 45
GPA: 3.9
The sum of the squares of the digits of a two digit number is 65. How  [#permalink]

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30 Dec 2019, 09:24
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84% (02:09) correct 16% (02:03) wrong based on 50 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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Joined: 04 Jan 2015
Posts: 3348
Re: The sum of the squares of the digits of a two digit number is 65. How  [#permalink]

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30 Dec 2019, 09:42
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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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Re: The sum of the squares of the digits of a two digit number is 65. How  [#permalink]

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30 Dec 2019, 10:29
1
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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Joined: 03 Jun 2019
Posts: 2502
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Re: The sum of the squares of the digits of a two digit number is 65. How  [#permalink]

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30 Dec 2019, 11:47
1
naju wrote:
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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Re: The sum of the squares of the digits of a two digit number is 65. How  [#permalink]

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24 Feb 2020, 01:20
naju wrote:
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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Re: The sum of the squares of the digits of a two digit number is 65. How   [#permalink] 24 Feb 2020, 01:20
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