The last 2 digits in mulplcn 122*123*125*127*129 will be....(do without calculator and quickly)..pls explain?

Answer 1

50.

Note that one of the numbers is an exact multiple of 25 and another is a multiple of 2. Thus your answer will be an exact multiple of 50. However, since only one number is divisible by two, and this number is not divisible by 4, your answer will not be a multiple of 4, and thus cannot be a multiple of 100 either. Since all multiples of 50 that are not multiples of 100 end in 50, you know that those are the last two digits of your number.

Answer 2

122 · 123 · 125 · 127 · 129
61 · 2 · 123 · 5 · 25 · 127 · 129
61 · 123 · 5 · 127 · 129 · 50

Since you will multiply the others by 50, then the only possible last 2 digits will be 50 or 00. when the product of the first digits is even, then the last 2 are 00; when the product of the first digits is odd, then the last 2 are 50.

Since all the other first digits are odd, then their product is also odd.
Therefore the last 2 digits are 5 and 0.

^_^

Answer 3

Just ignore anything which will give an answer of 100 or higher.
2*3*5*7>10, so just multiply the 1's place digits together--all other parts of the product (eg 20*2*3*5*7) will end up in the 100's place or higher.
3*9=27 and 5*7=35. So this will be 2*(27*35). 20*5 =100 and 7*30>100, and 20*30>100. So you are left with 2*(7*5)=70.

The others are answering the problem as it appears in the headline--the full problem appears only when you elect to answer it. Nice job, Yahoo!

Answer 4

product is divisible by 125 and 2. hence 250
but not by 500

hence last two digits would be 50

Answer 5

the product = 5^2*2^1*...
so the two last digits must be 50

Answer 6

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