Question

Jo is thinking of a positive integer less than 100. It is one less than a multiple of 8, and it is three less than a multiple of 7. What is the greatest possible integer Jo could be thinking of

Answer 1

Answer:

25！=1*2*3*4*5*6*...23*24*25

26= 2*13

28=2*14or 4*7

36=2*18or 3*12 or 4*9

56=7*8

All products above are inside 25! So they are all factors of 25!

But 58=2*29 and 29 is not inside 25! so it's not a factor of 25!

Step-by-step explanation:

hope it helped

Answer 2

Answer:

n=25

Step-by-step explanation:

Let n be the greatest possible integer Jo could be thinking of. We know n<100 and n=8k-1=7l-3 for some positive integers k and l. From this, we see that 7l=8k+2=2(4k+1), so $7l$ is a multiple of 14. List some multiples of 14, in decreasing order: 112, 98, 84, 70, .... Since n<100, 112 is too large, but 98 works: $7k=98\Rightarrow n=98-3=95=8(12)-1$. Thus,$n=\boxed{95}$.

Hope this helped! :)