What number has a remainder of 1 when divided by 6 or 7 and a remainder of 7 when divided by 8 or 11?

Answer 1

Call the number we are looking for x. Since the number has a remainder of 1 when divided by 6 or 7, x-1 is divisible by 6 and 7, which means that x-1 is divisible by 42. Similarly, x-7 is divisible by 88.

The distance between x-7 and x-1 is 6. Therefore, we are looking for a number which is divisible by 88 and, when added to 6, is divisible by 42 (since when we find this number we can find the number we were originally looking for).

So far we have that x-7 is divisible by 88 and that (x-7)+6 is divisible by 42. Since 42 is divisible by 6, (x-7)+6 is also divisible by 6. This means that x-7 is also divisible by 6. This means that, since x-7 is also divisible by 88, it must be divisible by 264.

Now the restrictions for (x-7)+6 are that it must be divisible by 264 and that (x-7)+6 must be divisible by 7. However, if (x-7)+6 is divisible by 7, then so is (x-7)-1.

At this point we only have to test at most 7 cases to find a number that works.

However, there is a way to deal with smaller numbers.

The remainder when 264 is divided by 7 is 5. (To do this you can use the divisibility rule for 7 (double the last digit and subtract the number from the number formed when you take all of the digits of the number other than the units digit) or you could say that, since 21 is a multiple of 7, 210 is a multiple of 7. Therefore, the remainder when 264 is divided by 7 is the same as the remainder when 264-210, or 54, is divided by 7. Since 49 is a multiple of seven, the remainder when 54 is divided by 7 is the same as the remainder then 54-49, or 5, is divided by 7, which is 5.) We want to find the number which when multiplied by 264, gives a remainder of 1 when divided by 7. In other words, we want to to find a multiple of 7 that is 1 less than a multiple of 5. Since all multiples of 5 end in 0 or 5, the number we are looking for must end in 4 or 9.

Going through the multiples of 7, we see that 7*2, or 14, ends in a 4. This is one less than 15, which is divisible by 5. 15/5 is 3, the number of times you have to multiply 264 to get a number which is one more than a multiple of 7 is 3.

Let's find the number. 264*3=792. To test, 792-1=791, which is 7*113. Remember that this is x-7. Therefore, to find x, all we have to do is add 7 to 792, which gives 799.

It's always a good idea to check unless you have a time constraint, so let's check.

To find the remainder when 799 is divided by 6, first notice that 666 is divisible by 6. Therefore, the remainder when 799 is divided by 6 is the same as the remainder when 799-666=133 is divided by 6. Next, note that 120 is divisible by 6. Now we have to find the remainder when 133-120=13 is divided by 6. Since 12 is a multiple of 6, we take 13-12 to find that 799 has a remainder of 1 when divided by 6. Check!

Now, for 7. Since 700 is divisible by 7 and 91 is divisible by 7, 791 must be divisible by 7. This is pretty close to 799, so let's see what happens when we add one more 7 to 791. We get 798, which is one less than 799, so the remainder when 799 is divided by 7 is 1 as well. Check!

Now, for 8. 800 is divisible by 8, so the remainder when 799 is divided by 8 is the same as when 799-800, or -1 is divided by 8. -1 is equivalent to 7 when dividing by 8, so the remainder is 7. Check!

Finally, we check 11. Remember the rule that, to multiply a 2-digit number whose digits sum to less than ten by 11, take the sum of the two digits, and put it in the middle. 7+2=9, so 792 is divisible by 11. 799-792=7. Check!

Since our number satisfies all of the requirements, the number 799 is an answer.

To find other answers, take the 3 we found earlier and add 7 as many times as you want to it. Now, multiply that by 264 and add 7. Any of these numbers should work because they are 7 more than a multiple of 264 and the number 6 more than 7 less than the number is a multiple of 7 (because we added a multiple of 7 to it, so the remainder when divided by 7 should stay the same).