finding the last two digits

bhumika.k.shah wrote:Find the last 2 digits
65*29*37*63*71*87*62

1. 70
2. 30
3. 10
4. 90
5.94

OA D

The last digit will be 0 (5 and 2 are factors)
and the tens digit will be the last digit of
13*29*37*63*87*31 (dividing 65 by 5 and 62 by2)

Which is the last digit of 3*9*7*3*7*1
which is the last digit of 27*21*7
which is the last digit of 7*7
which is 9

so option D

ajith wrote:

bhumika.k.shah wrote:Find the last 2 digits
65*29*37*63*71*87*62

1. 70
2. 30
3. 10
4. 90
5.94

OA D

The last digit will be 0 (5 and 2 are factors)
and the tens digit will be the last digit of
13*29*37*63*87*31 (dividing 65 by 5 and 62 by2)

Which is the last digit of 3*9*7*3*7*1
which is the last digit of 27*21*7
which is the last digit of 7*7
which is 9

so option D

That's harmonious, really!

genius!!!

ajith wrote:

bhumika.k.shah wrote:Find the last 2 digits
65*29*37*63*71*87*62

1. 70
2. 30
3. 10
4. 90
5.94

OA D

The last digit will be 0 (5 and 2 are factors)
and the tens digit will be the last digit of
13*29*37*63*87*31 (dividing 65 by 5 and 62 by2)

Which is the last digit of 3*9*7*3*7*1
which is the last digit of 27*21*7
which is the last digit of 7*7
which is 9

so option D

ajith,

why'd u divide 65 and 62 by 5 and 2 resp?

rahul.s wrote:ajith,

why'd u divide 65 and 62 by 5 and 2 resp?

a. If a number is divisible by 5 and 2, the last digit is 0
b. The second last digit is the last digit of the number divided by 5 and 2

Example say you want to know the last two digits of 5*2*13*7*11

a. if we can divide the num by 5&2 the last digit is 0
b. the second last digit is the last digit of 13*7*11 which is 1

so last two digits of num is 10

rahul.s wrote:ajith,

why'd u divide 65 and 62 by 5 and 2 resp?

probably to take the only possible 10 as factor, out of it