Find the greatest number which when divided by 322,162 and 402 leave a remainder 2 in each case?
Answers
Answered by
2
GCD(137–2,182–2,422–2)GCD(137–2,182–2,422–2)
=GCD(135,180,420)=GCD(135,180,420)
Note, because the remainder is 2 (and because not all the numbers are even), we can exclude 2 as a factor. Next, check prime 3. Sum of the digits of 135 is 9, which is divisible by 3. The sum of the digits of 180 is 9, which is divisible by 3. The sum of the digits of 420 is 6, which is divisible by 3. Note, only one is divisible by 9, therefore, there is only one factor of 3 in common. Thus, we can move to check if 5 is a factor. Indeed, all three numbers either end in a 0 or a 5, therefore they are all divisible by 5. Two of the numbers end in 0, but do not have a 0 or 5 as their next digit, meaning there is only one factor of 5. Now, either divide 135 by 5, to get 27 (note, dividing by 5 is the same as multiplying by 2, then dividing by 10, which is, shifting right by one digit), which is only divisible by 3, or divide 180 by 15 to get 12, which is only divisible by 2 or 3. Since all additional factors of 2 or 3 are excluded, the result is the 15 is the largest number that divides all three numbers and leaves a remainder of 2 in each case.
USE THIS METHOD TO GET YOUR ANSWER
HOPE IT HELPS
=GCD(135,180,420)=GCD(135,180,420)
Note, because the remainder is 2 (and because not all the numbers are even), we can exclude 2 as a factor. Next, check prime 3. Sum of the digits of 135 is 9, which is divisible by 3. The sum of the digits of 180 is 9, which is divisible by 3. The sum of the digits of 420 is 6, which is divisible by 3. Note, only one is divisible by 9, therefore, there is only one factor of 3 in common. Thus, we can move to check if 5 is a factor. Indeed, all three numbers either end in a 0 or a 5, therefore they are all divisible by 5. Two of the numbers end in 0, but do not have a 0 or 5 as their next digit, meaning there is only one factor of 5. Now, either divide 135 by 5, to get 27 (note, dividing by 5 is the same as multiplying by 2, then dividing by 10, which is, shifting right by one digit), which is only divisible by 3, or divide 180 by 15 to get 12, which is only divisible by 2 or 3. Since all additional factors of 2 or 3 are excluded, the result is the 15 is the largest number that divides all three numbers and leaves a remainder of 2 in each case.
USE THIS METHOD TO GET YOUR ANSWER
HOPE IT HELPS
Similar questions