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Number Theory Level 3

Find the sum of all positive integers x smaller than 100 such that x 2 − 81 x^{2} - 81 is a multiple of 100 greater than 0.


The answer is 191.

Kenneth Gravamen by Kenneth Gravamen , 23, Philippines

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1 solution

Let x = 10 a + b x = 10a + b for 1 ≤ a ≤ 9 1 \le a \le 9 and 0 ≤ b ≤ 9. 0 \le b \le 9. (Clearly x ≥ 10 x \ge 10 for x 2 − 81 > 0 x^{2} - 81 \gt 0 .)

Then we require that ( 10 a + b ) 2 ≡ 81 ( m o d 100 ) (10a + b)^{2} \equiv 81 \pmod{100}

⟹ 10 a 2 + 20 a b + b 2 ≡ 81 ( m o d 100 ) ⟹ 20 a b + b 2 ≡ 81 ( m o d 100 ) . \Longrightarrow 10a^{2} + 20ab + b^{2} \equiv 81 \pmod{100} \Longrightarrow 20ab + b^{2} \equiv 81 \pmod{100}.

Now since the units digit of 20 a b 20ab will always be 0 0 we will require that the units digit of b 2 b^{2} is 1. 1. This implies that b b must either be 1 1 or 9. 9.

If b = 1 b = 1 then we require that 20 a + 1 ≡ 81 ( m o d 100 ) ⟹ 20 a ≡ 80 ( m o d 100 ) 20a + 1 \equiv 81 \pmod{100} \Longrightarrow 20a \equiv 80 \pmod{100}

a ≡ 4 ( m o d 5 ) ⟹ a = 4 , 9 , a \equiv 4 \pmod{5} \Longrightarrow a = 4,9, which gives us the solution values 41 41 and 91. 91.

If b = 9 b = 9 then we require that 180 a + 81 ≡ 81 ( m o d 100 ) ⟹ 180 a ≡ 0 ( m o d 100 ) 180a + 81 \equiv 81 \pmod{100} \Longrightarrow 180a \equiv 0 \pmod{100}

⟹ 9 a ≡ 0 ( m o d 5 ) ⟹ a = 5 , \Longrightarrow 9a \equiv 0 \pmod{5} \Longrightarrow a = 5, which gives us the solution value 59. 59.

The sum of al possible solution values is then 41 + 59 + 91 = 191 . 41 + 59 + 91 = \boxed{191}.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

x^2_81=(x-9)(x+9) now if one of these factors=25 or 75 the other is not not divisible by four So one of the factors=50 or 100 giving the other being even So admissible values for x are 41,59 and 91(as x<100 we ignore x=109)

Des O Carroll - 5 years, 8 months ago

I use Diophantine.

Kenneth Gravamen - 5 years, 8 months ago

i dont know that much maths but cant we think like this that if x=9 , x^2 -81 will be 0, if x=10 x^2 -81=19,ifx=11, x^2-81taking like this we will get 19+40+63+88 =210

Aaysha Babu - 5 years, 7 months ago

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But x 2 − 81 x^{2} - 81 needs to be a positive multiple of 100 , 100, which is not the case for x = 10 , 11 , 12 , 13 , x = 10,11,12,13, and is only the case for x = 41 , 59 , 91 x = 41, 59, 91 for positive values of x x under 100. 100.

Brian Charlesworth - 5 years, 7 months ago

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sorry it was my mistake i did'nt read the question carefully thnks now i understand the solution

Aaysha Babu - 5 years, 7 months ago

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