Last 2 digits of a product of numbers

Multiply the single digit of all numbers - 3*4*6*2*3 = 432. 32 is the last 2 digits of the product.

@ papgust
The product of 23*84*96*12*123 is 273756672
the last two digit is 72
I feel Animan should give the Total problem not just a part of it.

papgust wrote:Multiply the single digit of all numbers - 3*4*6*2*3 = 432. 32 is the last 2 digits of the product.

I originally made this same error. When we care about the last two digits, we need to multiply the last two digits of every number, not just the last digit.

For example, the last two digits of 12*12 aren't 04 - they're 44; the last two digits of 20*6 aren't 0 - they're 20.

When I realized this, I gave up on this question since there's no quick way to do it; in other words, it's not a realistic GMAT question.

If we had the full question, with choices, we could provide useful tips; for example, if the last digit of every answer were different, then we could indeed just multiply the last digits together and find a match.

Without the full question and the choices, however, there's no sense spending time on this problem.

Sorry, actually I got this question in Internet where the options were given as 32, 72, 62, 51 and 52.

51 can be easily ruled out but I was thinking instead of hit and trial whether there is any proper logical method to deduce this. I've found out a link to help me getting the value of a number raised to a huge power but I can't deduce anything for this kind of problem

https://cat.learnhub.com/lesson/5288-fin ... -of-any-no

First time poster here.

Sorry for bumping this old post, but is there an easy way to solve this? We can obviously rule out 51, but the others all end in 2...!
I looked at the Veritas Prep method with remainders and it's a bit confusing/takes just as long as multiplying all the numbers out.

Any help would be much appreciated.