Question

Find the last two digits of the product 122 × 123 × 125 × 127 × 129?

Answer 1

Hello my friend....
Here is your answer.....
The product of these number is
122× 123× 125× 127× 129 =30730412250
So the last two digits of the product is 50.
Hope this help you ✌️ ✌️ ✌️....

Answer 2

Answer:

50

Step-by-step explanation:

Generalization

Let n = 120, then given product can be written as:

$= (n+2)\times (n+3)\times(n+5)\times(n+7)\times(n+9)\\=( n^2 +5n +6)\times(n^2+12n+35)\times(n+9)\\=(n^4+17n^3+95n^2+237n+210)\times(n+9)\\= n^5+26n^4+228n^3+1102n^2+2133n+1890\\\implies (n^3 +26n^2+228n +1102)n^2 +2133n +1890\\\implies K(n)n^2 + [2133n+1890]$

Now, notice that $n^2$ term will give 14400 and when multiplied by K(n) it still contains two zero in the end, which won't change the solution; so for last two digits we only look towards term 2133n + 1890.

Again, note that when 120(=n) multiplied by '3' (last digit of 2133); we have last two digits as 60.

when 60 gets added with 90 ( of the term 1890) we get 50 as last two digits.

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