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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, 4 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

\( 180,90,60,45,36,30,20 \)

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## Comments

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TopNewestLet 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

\( 180,90,60,45,36,30,20 \)

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which class r u in?

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