How many times 1 6 2 should appear under the square root sign for the following equation to be true:
1 6 2 + 1 6 2 + 1 6 2 + . . . + 1 6 2 + 1 6 2 + 1 6 2 = 1 6 1 6 ?
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Let the number of times 1 6 2 comes under the square root be x . Then, x × 1 6 2 = 1 6 1 6 ⇒ x × 1 6 2 = ( 1 6 1 6 ) 2 = 1 6 3 2 ⇒ x = 1 6 2 1 6 3 2 = 1 6 3 0 .
Let x be the number of times 16^2 should appear. We find x to be equal to 16^15. Squaring it we get 16^30
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Rewriting this equation:
1 6 2 + 1 6 2 + 1 6 2 + . . . + 1 6 2 + 1 6 2 + 1 6 2 = 1 6 1 6 becomes n ⋅ 1 6 2 = 1 6 1 6 .
Solving:
n ⋅ 1 6 2 = 1 6 1 6
1 6 n = 1 6 1 6
n = 1 6 1 5
n = 1 6 3 0