Is This Algebra or Number Theory?

Algebra Level 3

Find the value of n n , such that 1 + 2 + 3 + 4 + 5 + 6 + 7 + ⋯ + n 1+2+3+4+5+6+7+ \cdots + n is equal to the 3-digit integer, a a a ‾ \overline{aaa} .

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The answer is 36.

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Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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1 solution

Freddie Hand
Jan 28, 2017

a a a ‾ = a × 3 × 37 \overline{aaa}=a×3×37

Also, 1 + 2 + 3 + . . . + n = n ( n + 1 ) 2 1+2+3+...+n=\frac{n(n+1)}{2}

As n ( n + 1 ) 2 = a × 3 × 37 \frac{n(n+1)}{2}=a×3×37 and 37 is prime, then 37 ∣ n ( n + 1 ) 2 37|\frac{n(n+1)}{2}

Therefore, n = 36 n=36 or n = 37 n=37 .

Testing n = 36 n=36 , we get a = 6 a=6

Therefore, n = 36 n=36 .

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice observation that a a a ‾ = 100 a + 10 a + a = 111 a = 3 × 37 × a \overline{aaa} = 100a + 10a + a = 111a = 3 \times 37 \times a .

Brian Charlesworth - 4 years, 4 months ago

Nice solution. That's what i'm expecting! Thank you!

Fidel Simanjuntak - 4 years, 4 months ago

GG, well played

genis dude - 4 years, 4 months ago

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