What two-digit number is less than the sum of the square of

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What two-digit number is less than the sum of the square of [#permalink]

15 Nov 2012, 07:03

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What two-digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5?

(A) 95
(B) 99
(C) 26
(D) 73
(E) None of the Above

[Reveal] Spoiler: OA

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Re: What two-digit number [#permalink]

15 Nov 2012, 07:29

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vomhorizon wrote:

What two-digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5?

(A) 95
(B) 99
(C) 26
(D) 73
(E) None of the Above

We are told that the two-digit number exceeds doubled product of its digits by 5:

(10a+b)-2ab=5 --> 2a(5-b)-(5-b)=0 --> (5-b)(2a-1)=0 --> b=5 (a cannot equal to 1/2 since it must be an integer). The only answer choice with 5 as an units digit is A.

Check 95 for the first condition (to eliminate E), which says that the two-digit number is less than the sum of the square of its digits by 11: (9^2+5^2)-95=11. So, the answer is A.

Answer: A.

There is another number satisfying both conditions:

Substitute b=5 in (a^2+b^2)-(10a+b)=11 --> a^2-10a+9=0 --> a=9 or a=1. Therefore both 15 and 95 satisfy both conditions.

Hope it's clear.
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Re: What two-digit number [#permalink] 15 Nov 2012, 07:29

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What two-digit number is less than the sum of the square of

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What two-digit number is less than the sum of the square of

What two-digit number is less than the sum of the square of

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