When 19911 and 29311 are divided by a natural number 'n, the remainder obtained in both cases is the same, R another natural number. How many possible values can 'n' take?
Answers
Given :- When 19911 and 29311 are divided by a natural number 'N', the remainder obtained in both cases is the same, R another natural number. How many possible values can 'n' take ?
Answer :-
given that, 19911 and 29311 are divided by a natural number 'n, the remainder gets is R .
so,
→ 19911 = aN + R ------ Eqn.(1)
→ 29311 = bN + R ------- Eqn.(2)
subtracting Eqn.(1) from Eqn.(2),
→ 29311 - 19911 = (bN + R) - (aN + R)
→ 9400 = bN - aN + R - R
→ 9400 = N(b - a)
→ N(b - a) = 2³ * 5² * 47
then,
→ Total number of factors of N are = (3 + 1) * (2 + 1) * (1 + 1) = 4 * 3 * 2 = 24 .
Therefore, N can take 24 possible values .
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Answer:
Step-by-step explanation:sjsnsn