Problem Solving

Find the greatest 4 digit number which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively
a)9907 b)9903 c)9893 d) None of these


Find the greatest number , which will divide 215,167 and 135 so as to leave the same remainder in each case
a)64 b)32 c)24 d)16


What is the smallest number which when increased by 6 is divisible by 36,63 and 108?
a) 750 b)752 c)754 d)756

sukhman wrote:Find the greatest 4 digit number which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively
a)9907 b)9903 c)9893 d) None of these

Dividing a number by 10 generates a remainder equal to the units digit. Only B and C work.

9900 divides evenly by 11. So dividing 9903 by 11 generates a remainder of 3 and 9893 by 11 a reminder of 4.

We are down to C or none.

99 is multiple of 3 and 100 is a multiple of 5. So 9900 is a multiple of 3 x 5 = 15. 9893 is 7 less than 9900. so the remainder when 9893 is divided by 15 is 8. C Still works.

99 is a multiple of 11 and 100 is a multiple of 2. So 9900 is a multiple of 11 x 2 = 22. 9893 is 7 less than 9900. 22 - 7 = 15, which is the remainder when 9893 is divided by 22.

So the correct answer is C.

Find the greatest number that will divide 215,167 and 135 so as to leave the same remainder in each case
a)64 b)32 c)24 d)16

Scanning the numbers I see that 167 is just over 160 = 10 x 16. So the remainder generated by 167/16 is 7.

The factors 10 x 16 can be broken down to 5 x 2 x 16. So 5 x 32 = 160, and the remainder generated by 167/32 is 7.

24 = 3 x 8. 160 does not have 3 as a factor. So 24 cannot be a factor of 160 and 167/24 will not generate the remainder 7.

64 = 2 x 32. So 64 cannot be a factor of 160, or 5 x 32. So 167/64 will not generate a remainder 7.

So we are down to 16 and 32.

The multiple of 32 closest to 135 is 4 x 32 = 128. So 135/32 and 135/16 generate the same remainder 135 - 128 = 7.

215/16 generates 13 x 16 = 208 with remainder 7. 32 = 2 x 16. So, multiplying 16 by an odd number, 13, will not generate a multiple of 32. So 208 is not a multiple of 32. Therefore the remainder generated by 208/32 will not be 7.

So 16 is the greatest number that works and the correct answer is D.

What is the smallest number which when increased by 6 is divisible by 36,63 and 108?
a) 750 b)752 c)754 d)756

Scanning the numbers I see that 756 = (7 x 100) + (7 x 8) So 756 is divisible by 108.

Therefore the only number in the list that when increased by 6 is divisible by 108 is 750.

So the answer must be A.

I believe this type of question is tested in CAT exam BUT NOT in GMAT.

I haven't seen or heard about this type of questions in exam.



Experts/ Instructors...Do you agree that this type have very less or no chance of appearing in GMAT?

These are not GMAT like questions as all questions have 4 options instead of 5 and Also they require massive calculation if not tried with options unlike what GMAT expects

sukhman wrote:Find the greatest 4 digit number which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively
a)9907 b)9903 c)9893 d) None of these

Check options and
Option A is clearly out as when divided by 10 that doesn't leave remainder 3
Option B is ruled out as when divided by 15 that doesn't leave remainder 8 (Remainder =3)
Option C is the answer

Find the greatest number , which will divide 215,167 and 135 so as to leave the same remainder in each case
a)64 b)32 c)24 d)16

Check with Options again

What is the smallest number which when increased by 6 is divisible by 36,63 and 108?
a) 750 b)752 c)754 d)756

Check with Options again

sukhman wrote:Find the greatest 4 digit number which when divided by 10,11,15 and 22 leaves 3,4,8 and 15 as remainders respectively
a)9907 b)9903 c)9893 d) None of these

3 mod 10 means that it must end in 3: A or C.

9907 = 11*900 + 7, so this doesn't give us 4 mod 11: eliminate A.

9893 = 9900 - 15 + 8 and = 8800 + 110 - 22 + 15, so 9893 has all the properties we want. The next four digit number that = 3 mod 10 and = 4 mod 11 must be 110 greater than this (11*10), but there is no four digit number 110 greater than 9893.

sukhman wrote:Find the greatest number , which will divide 215,167 and 135 so as to leave the same remainder in each case
a)64 b)32 c)24 d)16

Let's call this number n. We're given that

215 = 167 mod n, or 48 = 0 mod n, so n is a factor of 48: eliminate A and B.

We're also given that

215 = 135 mod n, or 80 = 0 mod n, so n is a factor of 80: only D remains.

sukhman wrote:What is the smallest number which when increased by 6 is divisible by 36,63 and 108?
a) 750 b)752 c)754 d)756

x + 6 = 0 mod 36, so x = -6 = 30 mod 36

The only number given here with this property is 750 (= 36*20 + 30), so this must be the answer.

vishalwin wrote:I believe this type of question is tested in CAT exam BUT NOT in GMAT.

I haven't seen or heard about this type of questions in exam.



Experts/ Instructors...Do you agree that this type have very less or no chance of appearing in GMAT?

These all look like questions from the Indian CAT, which assumes a much higher level of mathematical development than the GMAT does. Two of these questions are pretty easy with modular arithmetic, but the first one is rough: it isn't *that* hard to show that C satisfies the conditions, but it's hard to show that there is no GREATER four-digit number that satisfies the conditions without some fluency in the Chinese Remainder Theorem (or equivalent), which is way beyond the GMAT. (And in general, when I tried some old Indian CATs, I found that the problems for which "None of these" was an answer were very difficult to finish in time.)

... and just to tack on to my previous response, none of the other solutions above rule out (or even address) "None of these" in the first problem, showing just how daunting it is. (We're used to the GMAT, in which you don't ever have such a challenging answer to contemplate.)

One other way of doing so is using the aforementioned Chinese Remainder Theorem to find the root solution, then adding the LCM of the four divisors to the root.

Let's start with n = 3 mod 10 and n = 4 mod 11. The smallest such number is 103, which isn't too elusive. Any other solutions will be of the form 10*11*(some integer) + 103, so our set is 103, 213, 323, etc.

Notice too that n = 15 mod 22 is a redundancy: since 10*11*(some integer) is always divisible by 22, we only have to consider 103 mod 22, which itself (of course) ends up being 15. So we can ignore this.

That leaves only n = 8 mod 15. 103 and 213 don't have this property, but 323 does, so that's our root solution.

Every other solution must be of the form 323 + LCM(10,11,15)*(some integer), or 323 + 330*(some integer), or 323 + 330m, where m is some integer.

Now we seek the largest four digit integer of the form 330m + 323. 330*30 is too big, but 330*29 comes in below 10,000. 330*29 + 323 is ... (drumroll please!) 9893!

If none of this makes a lick of sense and doesn't interest you, IGNORE IT: it has no bearing on the GMAT, nor (quite likely) on anything you'll encounter in the rest of your life on earth.