Hey, I thought about another approach and I got 1 instead of 5. Can you please explain where I am wrong?
For my example, I will use smaller numbers.
3,1555,7
(3*1555*7)/10
Now I can reduce by 5 (because of the 1555)
1555/10 = 331/2
So I can see the result after reducing as (3*331*7)/2
And now the reminder must be zero or one, and because all the numbers are odd, the reminder must be 1.
Where is my mistake in my logic?
GreenlightTestPrep wrote:
Carcass wrote:
What is the remainder when 1551*1553*1555*1557*1559 is divided by 10?
(A) 0
(B) 2
(C) 5
(D) 7
(E) 9
Important property: The remainder when positive integer N is divided by 10 is equal to the units digit of 10So, the question is really asking us to find the units digit of the product (1551)(1553)(1555)(1557)(1559)Since all five numbers in the product are ODD, the product must be ODD, which means the units digit must be odd.
So, we can eliminate choices A and B
Next, since 1555 is a multiple of 5, we know that the entire product must be a multiple of 5.
So, the units digit of (1551)(1553)(1555)(1557(1559) is either 5 or 0, which means
the correct answer is C (since we've already eliminated choice A)