Question

Show that

3^{n} \times 4^{m}

cannot end with the digit

0

or

5

for any natural numbers

' n'

and

' m'

.

Accepted Answer

If any number end with digit $0$ , it should be divisible by $10$ or in other words, it will also be divisible by $2$ and $5$ as $10 = 2 \times 5$ .

And, if any number end with digit $5$ , it should be divisible by $5$ .

But here in the given number $3^{n} \times 4^{m} = 3^{n} \times {(2 \times 2)}^{m}$ , we observe that $2$ and $3$ are the only factors present for any natural number $n$ and $m .$

It can be observed that $5$ is not in the prime factorization of $3^{n} \times 4^{m}$ .

Hence, for any value of $n$ and $m$ , $3^{n} \times 4^{m}$ will not be divisible by $5$ .

$∴ 3^{n} \times 4^{m}$ cannot end with the digit $0$ or $5$ for any natural number $n$ and $m$ .

Hence, proved.

Telangana Board Solutions for Chapter: Real Numbers, Exercise 7: Try This

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Real Numbers, Exercise 7: Try This

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