Question

Qᴜᴇ - Let m = 20¹⁵ and n = 15²⁰. Let A denote the least common divisor of m and n. Find the number of trailing zeroes of A × B.​


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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Product of LCM and HCF has been used. We are given the two numbers with their LCM that is Least Common Multiple and HCF that is Greatest Common Divisor or Highest Commin Factor as A and B respectively. We need to find the trailing zeroes of A × B. This can be done by obtaining a equation of A × B where there is a term like 10ⁿ . This n will show the number of trailing zeroes.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{LCM\;\times\;HCF\;=\;\bf{Product\;of\;Two\;Numbers}}}$

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★ Solution :-

Given,

» First Number, m = 20¹⁵

» Second Number, n = 15²⁰

» LCM of m and n = A

» HCF of m and n = B

We know that,

$\\\;\;\sf{:\Longrightarrow\;\;LCM\;\times\;HCF\;=\;\bf{Product\;of\;Two\;Numbers}}$

By applying the values here, we get,

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{20^{15}\;\times\;15^{20}}}$

On splitting the terms, we get,

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{(5\;\times\;4)^{15}\;\times\;(3\;\times\;5)^{20}}}$

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{(5\;\times\;2\;\times\;2)^{15}\;\times\;(3\;\times\;5)^{20}}}$

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{(5\;\times\;2^{2})^{15}\;\times\;(3\;\times\;5)^{20}}}$

Now separating the exponents with bases, we get,

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{15}\;\times\;2^{2\;\times\;15}\;\times\;3^{20}\;\times\;5^{20}}}$

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{15}\;\times\;2^{30}\;\times\;3^{20}\;\times\;5^{20}}}$

Further combining the same bases, we get,

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{15\;+\;20}\;\times\;2^{30}\;\times\;3^{20}}}$

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{35}\;\times\;2^{30}\;\times\;3^{20}}}$

This can be written as,

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{30}\;\times\;5^{5}\;\times\;2^{30}\;\times\;3^{20}}}$

$\\\;\;\sf{:\Longrightarrow\;\;A\;\times\;B\;=\;\bf{5^{5}\;\times\;(5\;\times\;2)^{30}\;\times\;3^{20}}}$

$\\\;\;\bf{:\mapsto\;\;A\;\times\;B\;=\;\bf{5^{5}\;\times\;\underline{10^{30}}\;\times\;3^{20}}}$

Since, here the highest power of 10 is 30 that is 10³⁰.

This means the trailing zeroes of A × B will be 30 zeroes.

$\\\;\underline{\boxed{\tt{Thus,\;\;no.\;of\;\;trailing\;\;zeroes\;\;of\;\;A\;\times\;B\;=\;\bf{30\;\;zeroes}}}}$

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★ Laws of Exponents :-

$\\\;\sf{\leadsto\;\;a^{m}\;\times\;a^{n}\;=\;a^{m\;+\;n}}$

$\\\;\sf{\leadsto\;\;(a^{m})^{n}\;=\;a^{m\;\times\;n}}$

$\\\;\sf{\leadsto\;\;(ab)^{n}\;=\;a^{n}\;\times\;b^{n}}$

$\\\;\sf{\leadsto\;\;\dfrac{a^{m}}{a^{n}}\;=\;a^{m\;-\;n}}$

$\\\;\sf{\leadsto\;\;a^{1/2}\;=\;\sqrt{a}}$

$\\\;\sf{\leadsto\;\;a^{0}\;=\;1}$