When a natural number ‘N’ is divided by a positive integer, we get 19 as a remainder. But when 10 times of same number ‘N’ is divided by the same integer we get 10 as a remainder. Then how many such integers are possible?

Note by Kshitij Johary
4 years, 9 months ago

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Let the integer be x, Then,A\q

$$N \equiv 19 \pmod{x}$$ - (1)

$$10N \equiv 10 \pmod{x}$$ - (2)

From (1),

$$10N \equiv 190 \pmod{x}$$

$$190 \equiv 10 \pmod{x}$$

$$180 \equiv 0 \pmod{x}$$

From (1) we have $$19 < x$$

x may be any divisor of 180 greater than 19 which are

$$180,90,60,45,36,30,20$$

- 4 years, 9 months ago

which class r u in?

- 4 years, 1 month ago