If-a-b-N-and-a-b-100-a-6-b-8-how-many-digits-can-the-product-have-at-most-Answer-17-

Question Number 174144 by Shrinava last updated on 25/Jul/22

If a, b \in N and a \cdot b = 100

a^{6} \cdot b^{8} how many digits can the product

have at most ?

Answer : 17

Answered by mr W last updated on 26/Jul/22

i f a n a t u r a l n u m b e r N h a s n d i g i t s, i t

m e a n s

10^{n - 1} ⩽ N < 10^{n}

\log (10^{n - 1}) ⩽ \log (N) < \log 10^{n}

n - 1 ⩽ \log (N) < n

\log (N) < n ⩽ \log (N) + 1

\Rightarrow n = ⌊ \log (N) ⌋ + 1

N = a^{6} b^{8} = {(\frac{100}{b})}^{6} b^{8} = 10^{12} b^{2}

N_{m a x} = 10^{12} \times {(100)}^{2} = 10^{16}

n = ⌊ \log N_{m a x} ⌋ + 1

= ⌊ \log 10^{16} ⌋ + 1

= ⌊ 16 ⌋ + 1

= 17

Commented by Shrinava last updated on 06/Aug/22

thank you dear professor