Do We Multiply Or Add?

Algebra Level 1

a 1 × a 2 × a 3 × × a n = a 276 \large a^1 \times a^2 \times a^3 \times \cdots \times a^n = a^{276}

Find the value of n n satisfying the equation above.

23 24 25 26

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2 solutions

Hung Woei Neoh
Jun 1, 2016

Relevant wiki: Sum of n, n², or n³

a 1 × a 2 × a 3 × × a n = a 276 a 1 + 2 + 3 + + n = a 276 1 + 2 + 3 + + n = 276 i = 1 n i = 276 n ( n + 1 ) 2 = 276 n 2 + n = 552 n 2 + n 552 = 0 ( n + 24 ) ( n 23 ) = 0 n = 24 , 23 a^1 \times a^2 \times a^3 \times \ldots \times a^n = a^{276}\\ a^{1+2+3+\ldots+n} = a^{276}\\ 1+2+3+\ldots+n = 276\\ \displaystyle \sum_{i=1}^n i = 276\\ \dfrac{n(n+1)}{2} = 276\\ n^2+n = 552\\ n^2 + n -552=0\\ (n+24)(n-23) = 0\\ n=-24,\;23

We know that n > 0 n > 0 , therefore the answer is n = 23 n=\boxed{23}

That is absolutely correct.

Hana Wehbi - 5 years ago

We first note that, a m × a n = a m + n \large a^m\times a^n = a^{m+n}

a 1 × a 2 × × . . . × a 23 \large \implies a^1\times a^2\times\times...\times a^{23}

= a n = 1 23 n = a 23 × 24 2 = a 276 \large = a^{\displaystyle \sum_{n=1}^{23} n } = a^{\frac{23\times24}{2}} = a^{276}

n = 23 \large \implies n= \boxed{23}

There is no doubt that you are 100% correct.

Hana Wehbi - 5 years ago

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