Numbers - tough questions

[gmatrant]

1. If 1<=k<=40, how many prime numbers are there which are of the form 5k+1?

Ans : 10

2.A three digit number of the form xyz is such that the numbers equals x!+y!+z!, find the difference of the number formed by reversing its digits and the number.

Ans : 396

3. A number when divided by a certain divisor leave remainders 238 and 342 respectively. When the sum of the two numbers is divided by the same divisor, the remainder is 156. Find the divisor/

Ans : 424

4.When the number 5,7,11 divide a multiple of 17, the remainders left are respectively 4,6,10. Which multiples of 17 gives the least number that satisfies the given condition?
Ans 317th

[tendays2go]

(1)Ans: 11
Number of the form 5K+1 is prime therefore should be odd as well.
therefore, it would be valid only for even values of K
for K=2 => 11 that is prime
for K =4 => 21 , not prime
for K=6 => 31 that is prime
for K =8 => 41 prime
similarly for K= 40 => 201 not a prime

so, we need to find out the prime numbers of the form 10n+1
where 1<= n <= 20
i.e. 11,31,41,61,71,101,131,151,161,181,191

(2)xyz = 100x +10y +z = x! + y! +z!
also zyx = 100z +10y+x
thus, zyx-xyz = 99(z-x)

also since it is a three digit number so xyz <999
also 6! = 720 and 7! = 5040
thus, x,y,z = {0,1,...6}
Also the biggest digit has to be either 5/6 to make xyz a 3 digit number
as it is sum of factorials of 3 digits too.
taking the largest digit as 6 need the number to be starting in 700s
thus largest digit is 5 i.e. z =5

by all this logic, number comes to be 145
thus, difference is: 99(5-1) = 396

[4meonly]

taking the largest digit as 6 need the number to be starting in 700s
thus largest digit is 5 i.e. z =5

Can not understand the logic

FROM: taking the largest digit as 6 need the number to be starting in 700s
TO: thus largest digit is 5 i.e. z =5

how you reasoned that the largest digit is 5?

[Ian Stewart]

(1)Ans: 11
Number of the form 5K+1 is prime therefore should be odd as well.
therefore, it would be valid only for even values of K
for K=2 => 11 that is prime
for K =4 => 21 , not prime
for K=6 => 31 that is prime
for K =8 => 41 prime
similarly for K= 40 => 201 not a prime

so, we need to find out the prime numbers of the form 10n+1
where 1<= n <= 20
i.e. 11,31,41,61,71,101,131,151,161,181,191

The logic is good, but 161 is not a prime; it is equal to 7 times 23. The other ten numbers are prime, however.

[Ian Stewart]

taking the largest digit as 6 need the number to be starting in 700s
thus largest digit is 5 i.e. z =5

Can not understand the logic

FROM: taking the largest digit as 6 need the number to be starting in 700s
TO: thus largest digit is 5 i.e. z =5

how you reasoned that the largest digit is 5?

The logic here is good, though it's a time consuming question- one must consider a few cases. tendays is saying that if the number contained the digit 7 (or 8 or 9), then the number would need to be larger than 7! (since the number equals x! + y! + z!). But if the number were larger than 7!, it would be at least a four-digit number. So we know the number cannot contain the digit 7, 8 or 9. But if the number contained the digit 6, it would need to be larger than 6! = 720, and would therefore need to contain the digit 7 (or 8, or 9). We just showed that's impossible. So the largest digit must be a 5; it cannot be a 4, because we could only get a two-digit number if the largest digit is a 4 (because 4! = 24).

[gmatrant]

Thanks a lot folks for your replies...
Can you help me with Q 3 & 4 as well.

As for question 1, i solved the same way tendays solved it, but was wondering if there was a easier way. Like for example it would be tough to say 161 is a multiple of 23*7, unless we calculate for each number i guess we might get to the wrong answer. So is this the best way to solve this question.

[lunarpower]

#3

before you solve this problem - which strikes me as something that's not likely to turn up on the exam - you should realize what happens with remainders when you add numbers.
specifically:
you can add remainders, but you have to "roll over" the remainder if it gets bigger than the divisor.

let's say you're dividing by, i don't know, um, 26.

let's say x divided by 26 gives a remainder of 8, and y divided by 26 gives a remainder of 11.
then (x + y) divided by 26 will give a remainder of 8 + 11 = 19.
this is still less than 26, so, no problem.

now let's say x divided by 26 gives a remainder of 18, and y divided by 26 gives a remainder of 21.
then 18 + 21 = 39.
but that's too big - if you have a "leftover" 39, then you can extract another 26, leaving a remainder of 13.

that's how it works.

--

SO:

in problem #3, adding the remainders gives 238 + 342 = 580.
but note that you "rolled over" back to 156.
therefore, you've subtracted 580 - 156 = 424, so 424 must be the divisor.
(note that the divisor can't be a smaller factor of 424 - it must be 424 itself - because 342, as a legitimate remainder, is smaller than the divisor)

[lunarpower]

4.When the number 5,7,11 divide a multiple of 17, the remainders left are respectively 4,6,10. Which multiples of 17 gives the least number that satisfies the given condition?
[/quote]

let's let the multiple of 17 be called "N".

since N gives a remainder of 4 upon division by 5, this means that N + 1 is a multiple of 5.

since N gives a remainder of 6 upon division by 7, this means that N + 1 is a multiple of 7.

since N gives a remainder of 10 upon division by 11, this means that N + 1 is a multiple of 11.

therefore, N + 1 is a multiple of 5, 7, and 11. since those are all prime, their lcm is 5 x 7 x 11 = 385.
therefore, you need N so that N is a multiple of 17, and N + 1 is a multiple of 385.
or, alternatively, you need Q so that Q is a multiple of 385 and Q - 1 is a multiple of 17. (here "Q" is the same as "N + 1")

i can think of a few ways to do this, but all are well beyond the purview of the gmat, and all would take way more than the allotted time. there is probably a brilliant, elegant solution that evades me.
anybody?

[Ian Stewart]

or, alternatively, you need Q so that Q is a multiple of 385 and Q - 1 is a multiple of 17. (here "Q" is the same as "N + 1")

i can think of a few ways to do this, but all are well beyond the purview of the gmat, and all would take way more than the allotted time. there is probably a brilliant, elegant solution that evades me.
anybody?

We're in territory here that's way beyond the GMAT, but with a bit of modular arithmetic, this can be completed fairly quickly. Anyone taking the GMAT- you can safely ignore what I'm writing below!

We know Q = 385k. We also know that the remainder is 1 when Q is divided by 17. Using modular arithmetic:

385k ~ 1 (mod 17)
11k ~ 1 (mod 17)

Notice:

11*3 = 33 ~ -1 (mod 17)

so:

11*(-3) = -33 ~ 1 (mod 17)

Thus k = -3 is one possible value of k, as is -3+17 = 14. Since we want the smallest k > 0, Q = 14*385, from which Q-1 can be found.

While I think some of the questions in the original post are interesting, none of them are much like real GMAT questions. The numbers are too awkward, and in some cases the techniques one needs go beyond what is expected of the GMAT test-taker.

[stop@800]

I enjoy solving these kind of questions.

Where can I find more tough questions like these?