Question

v. Find the smallest 5-digit number which when divided by 32,484 and 176,
leaves a remainder of 2 in each case

please explain by prosess​

Answer 1

Answer:

Should be 10,169 …… assuming one means divisible by each of 7, 11 and 21 a whole integer number of occasions. Otherwise 10,000 is the lowest 5 digit number, divisible by each of the starting numbers (7,11 & 21) and still leavlng a residual of 5 ….. plus a bit more.

10,169 is calculated by first finding the lowest common multiple (LCM) of start numbers 7x11x21 = 231 (or 11x21 since 21 is already 3x7).

Since 231 is less than our 5 digit quest … 10,000 ÷ 231 = 43+, we multiply by the next highest integer (i.e. 44) to get 231x44 = 10,164

Now to have a residual of 5 we simply add+5 to get the answer of 10,169.

A similar, supposedly logical, approach might yield a higher, but incorrect answer of 11,324 found by taking the product of start numbers (7,11 & 21) i.e. 1,617 and reasoning that 10,000 ÷ 1,617 =<7 …… Concluding, incorrectly, that 7x1,617 +5 (desired residual) or 11,324 was the solution. It is an answer meeting most criteria but it's not the lowest integer - using the LCM method is a more elegant approach.