What number leaves a remainder of 8 when divided by 9 a remainder of 7 when divided by 8 a remainder of 6 when divided by 7 a remainder of 5 when divided by 6 a remainder of 5 when divided by 5?

Answer 1

Let's call this number x and the quotients qi of x by i

So the statements are

x/9=q9+8

x/8=q8+7

x/7=q7+6

x/6=q6+5

x/5=q5+5

Look at the last one : x/5=q5+5, if one divides a number by 5, the remainder is either 0, 1, 2, 3 or 4, so there's no such a number

So I assume that the last statement should be x/5=q5+4

Let's denote ki=qi+1

x/9=q9+8 or x=9(q9+1)-1=9k9-1

x/8=q8+7 or x=8(q8+1)-1=8k8-1

x/7=q7+6 or x=7(q7+1)-1=7k7-1

x/6=q6+5 or x=6(q6+1)-1=6k6-1

x/5=q5+4 or x=5(q5+1)-1=5k5-1

so x+1=9k9=8k8=7k7=6k6=5k5

Then x+1 is a multiple of 9,8,7,6,5 so a multiple of their LCM

As 9 and 6 are not coprime and 8 and 6 are not coprime

LCM 9,8,7,6,5 is 9x8x7x6x5/(3x2) = 2520

then x+1=2520k with k in N

or x=2520k-1