JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 20)
Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number (10)10 . (11)11 . (13)13 is equal to __________.
Answer
924
Explanation
N = 210 $$\times$$ 510 $$\times$$ 1111 $$\times$$ 1313
Now, power of 2 must be zero,
power of 5 can be anything,
power of 13 can be anything
But, power of 11 should be even.
So, required number of divisors is
1 $$\times$$ 11 $$\times$$ 14 $$\times$$ 6 = 924
Now, power of 2 must be zero,
power of 5 can be anything,
power of 13 can be anything
But, power of 11 should be even.
So, required number of divisors is
1 $$\times$$ 11 $$\times$$ 14 $$\times$$ 6 = 924
Comments (0)
