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A four digit number is formed using $2,3,5,7$ or $9$ without repeating any of the digits. How many four digit number are there if each number has a remainder of $2$ when divided by either $3$ or $5$?

Hi, for my personal understanding, I understand that the last number for the four digit should be $2,5$ or $7$ since it must has remainder of $2$ when divided by $5$ or $3$. So the last number will contain three choices? I am not sure whether the answer should be like $(4)(3)(2)(3)$?

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  • $\begingroup$ If the last number is $5$, it will have a remainder of $0$ (not $2$) when divided by $5$. $\endgroup$
    – vadim123
    Nov 23, 2015 at 1:33
  • $\begingroup$ ^ right. So the last digit has to be $2$ or $7$. For the remainder being $2$ when divided by $3$, use the fact that a number is divisible by $3$ if and only if its digits in base $10$ sum to a multiple of $3$. $\endgroup$
    – Ethan Alwaise
    Nov 23, 2015 at 1:34
  • $\begingroup$ Ifhe sum of the digits will add up to 2 more than a multiple of 3 it will have a remainder of 2 when divided by 3. $\endgroup$
    – fleablood
    Nov 23, 2015 at 1:35
  • $\begingroup$ so the final calculation should be (4)(3)(2)(2)? since the last digit can only be either 2 or 7? but how about 7? if the number 7 is divided by 3, then the remainder would be left 1. $\endgroup$
    – hk23214
    Nov 23, 2015 at 1:41
  • $\begingroup$ 7 is only the last digit. You must add all the digits to see what the remainder divided by 3 is. $\endgroup$
    – fleablood
    Nov 23, 2015 at 1:45

1 Answer 1

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Let the number be abcd.

1) abcd divided by 3 has remainder 2.

So abcd = 1000a + 100b + 10c + d = 999a + 99b + 9c +a + b + c +d has remainder 2 so a + b + c + d has remainder 2.

2 + 3 + 5 + 7 + 9 = 26 = 24 + 2 has remainder 2.

If we subtract 3 or 9 from 26 we will have a remainder of 2. If we subtract 2, 5, or 7 from 26 we will not.

So the four digits are either 2, 3, 5 and 7 or 2, 5,7 and 9. Or in other words the four digits will contain 2, 5, and 7 and either one or the other of 3 or 9. We will call the four digits 2, 5, 7 and x which can be either 3 or 9.

2) abcd divided by 5 has remainder 2.

So abcd = 10*abc +d will have remainder 2 so d will have remainder 2.

2 and 7 are the only possible choices for d.

So there are 2 possible choices (2 or 7) for d. There are 3 possible choices for a (Either 5 or x or whichever wasn't chosen for d). There are 2 possible choices for b (One of the 2 that wasn't chosen for a or d, or the other one of the 2 that wasn't chosen for a or d). This is 1 possible choice for c (whatever is left). This is 2*3*2*1 = 12 possible numbers:

2x57; 25x7; x257; x527; 52x7; 5x27; x572; ....etc.

For each of these 12 possibilities x can be either 3 or 9 so there are 12*2 = 24 possible options.

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  • $\begingroup$ Hi. Thanks a lot! But I have some doubts that why we need to add abcd to determine the remainder ? Besides, {2,5,7,9} of {2,3,5,7} stands for what? and why there will be {2,5,7,x}? $\endgroup$
    – hk23214
    Nov 23, 2015 at 1:51
  • $\begingroup$ If you add the digits of a multiple of three the result is a multiple of 3. If the remainder is one higher or lower the remainder is one higher or lower. There are many places this is proven but in the case of 2 digits. 10a + b = (9 + 1)a + b = 9a + a + b. If you divide by 3 a+b and 10a + b will have the same remainder. This can be extended to indefinate digits. $\endgroup$
    – fleablood
    Nov 23, 2015 at 2:51
  • $\begingroup$ {2,5,7,9} and {2,3,5,7} are the two possible choices for the four digits regardless of order. 2579, 2795, etc will all have remainder 2 when divided by 3 as 2 + 5 + 7 + 9 = 23 and 2 + 3 = 5 has remainder of 2. The same is true for any 2357, 3275, 7532, etc. No other set of 4 digits will work. So... the set of possible choices for a,b,c,d are either the set 2,3,5,7 or 2,5,7,9 But that's just the choice of digits. The order can be anything. $\endgroup$
    – fleablood
    Nov 23, 2015 at 2:56
  • $\begingroup$ So there are two conditions for the number to have a remainder of 2 when divided by 3 or by 5. 1) the last digit is either 7 or 2. 2) The 4 digits are either 2,3,5,7 or 2,5,7,9. But now we need to calculate how many possibilities there are. The number of possibilities using 2,3,5,7 will be the same as using the numbers 2,5,7,9. So I said let x = 3 or 9. How many ways are there for the digits 2,5,7,x? (where x is either 3 or 9). $\endgroup$
    – fleablood
    Nov 23, 2015 at 3:00
  • $\begingroup$ The answer is that 2*3*\2*1 as there are 2 choices for d (2 or 7), 3 choices for a (evertyhing by d), 2 choices for b (everything but d and a) and 1 choice for c (whatevers left). That's 12 choices for either digits 2,3,5,7 or 2,5,7,9. As there are 2 possible sets of digits and 12 options for each of those sets, there are 24 total possibilities. $\endgroup$
    – fleablood
    Nov 23, 2015 at 3:04

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