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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  [#permalink]

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15 Nov 2012, 07:03
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45% (medium)

Question Stats:

68% (02:05) correct 32% (02:18) wrong based on 241 sessions

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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

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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.

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 is less than the sum of the square of  [#permalink]

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03 Sep 2014, 01:08
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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

Let the digits be x and y. The number would be 10x + y.
We are given that 2xy + 5 = 10x +y = x^2 y^2 -11
Thus 2xy +5 = x^2 + y^2 - 11
x^2 + y^2 -2xy = 16
(x-y)^2 = 16
(x-y) = 4 or -4

Substituting the values of (x-y) in the equation 2xy +5 = 10x + y
x comes out to be 1 or 9... thus the two numbers can be 15 or 95
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What two-digit number is less than the sum of the square of  [#permalink]

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21 Dec 2017, 18:40
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

Arithmetically, the operations on the given numbers are easy.

And there are two conditions. It's likely that the first one will eliminate at least one choice.

Start with just the first condition: the number is "less than the sum of the square of its digits by 11."

Translate: Square each digit. Add the results. Subtract 11. That should be the number.

(A) 95
$$9^2 = 81$$
$$5^2 = 25$$
Sum is 106. (196 - 11 ) = 95 KEEP

(B) 99
$$9^2 = 81$$
$$9^2 = 81$$
Sum is 162. STOP. Not even close to a difference of 11. REJECT

(C) 26
$$2^2 = 4$$
$$6^2 = 36$$
Sum is 40. (40 - 11) = 29 REJECT

(D) 73
$$7^2 = 49$$
$$3^2 = 6$$
Sum is 55. STOP. 55 < 73, let alone (55 - 11). REJECT

(E) None of the Above.
KEEP FOR NOW

One number, 95, satisfies first condition.

Second condition? It "must exceed [the digits'] doubled product by 5."
90 + 5 = 95. KEEP

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

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