Q11 of 39 Page 6

Show that 12n cannot end with the digit 0 or 5 for any natural number n.

If any number ends with the digit 0 or 5, it is always divisible by 5. If 12n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5.


12 = 2 × 2 × 3 = 22 × 3


⇒ 12n = (22 × 3)n = 22n × 3n


Since its prime factorisation does not contain 5.


Hence, 12n cannot end with the digit 0 or 5 for any natural number n.


More from this chapter

All 39 →
9

Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.

 10

Prove that is irrational.

 12

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45 cm, respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?

 13

Write the denominator of rational number in the form 2m × 5n, where m, n are non-negative integers. Hence, write its decimal expansion, without actual division.