Permutation and Combination

Question

How many 3 digit numbers can be formed from the digits 2 3 5 6 7 which are divisible by 3 (repetition allowed)

Archit3110 · Answer

divisiblity by 3 ; sum of digits should be divisible by 3
23567 following cases possible

222; 1 way
237; 6 way
267; 6 way
225; 3 way
336; 3 way
366; 3 way
333; 1 way
555;1 way
567; 6 way
666;1 way
777; 1 way

total ways ; 31

Kinshook · Answer

Asked: How many 3 digit numbers can be formed from the digits 2 3 5 6 7 which are divisible by 3 (repetition allowed)

For a number to be divisible by 3, its sum of digits must be divisible by 3.

Combinations 222:1,333:1,555:1,666:1,777:1,336:3,366:3,267:6,237:6,225:3,255:3 are divisible by 3. Total 1*5+3*4+6*2 = 5+12+12 = 29 such numbers can be formed.

AndrewN · Answer

I got a larger answer than the two previous posters did.  My answer is 41.  Consider:

Any time the three digits are the same, there will be one unique arrangement (3!/3!)  Any time two digits are the same, there will be three unique arrangements (3!/2!)   Any time each digit is unique, there will be six unique arrangements (3!)

There is exactly one way to achieve the least sum of 6: 2 + 2 + 2   There are two ways to achieve a sum of 9: any combination of 2-2-5 or the unique 3-3-3   There are several ways to achieve a sum of 12: any combination of 2-3-7, 2-5-5, or 3-3-6   There are several ways to achieve a sum of 15: any combination of 2-6-7, 3-5-7, 3-6-6, or the unique 5-5-5   There are two ways to achieve a sum of 18: any combination of 5-6-7 or the unique 6-6-6   Finally, there is exactly one way to achieve the greatest sum of 21: 7 + 7 + 7

Putting the information from the two lists together, we get

222 * 1
225 * 3
237 * 6
255 * 3
267 * 6
-------
333 * 1
336 * 3
357 * 6
366 * 3
-------
555 * 1
567 * 6
-------
666 * 1
-------
777 * 1

Thus, there are 41 valid combinations of the digits 2, 3, 5, 6, and 7 that can make up a three-digit number that is divisible by 3.

- Andrew

IanStewart · Answer

I also got 41:

• if the three digits are all the same, the number is always divisible by 3, for five possible numbers
• if two digits are the same and the other different, the digits can be, in some order, 225, 255, 336 or 366, and in each case we can arrange the three digits in three ways, for a further 12 possible numbers
• if the three digits are different, they can be, in some order, 237, 267, 357, or 567, and in each case we can arrange the three digits in 3! = 6 ways, for a further 24 possible numbers

and adding we have 41 possible numbers in total.

This is not, however, the kind of thing the GMAT is likely to ask, because to solve you just need to tediously enumerate a lot of cases, and GMAT questions aren't like that.

AndrewN · Answer

Hello, Ian. I agree with the point about the question not resembling one that would appear on the GMATTM, and for the very reason you stated. It was a fun exercise for me, one whose fundamentals I can appreciate, but that is all. I appreciate your taking the time to respond and providing not only your answer, but also the rationale behind that answer. (I would be willing to bet that I labored a bit longer than you on the problem, and I confess that I had originally worked it out with 4 included, only to catch myself at the end and eliminate those combinations.)

- Andrew

Permutation and Combination

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards