54$\times$10 = 540 and 45$\times$10 = 450. The difference between 540 and 450 is 90, which is much smaller than 198. So the number must have been much greater than 10.

54$\times$20 = 1080 and 45$\times$20 = 900. The difference between 1080 and 900 is 180, which is a bit smaller than 198. So the number must have been a bit more than 20.

54$\times$21 = 1134 and 45$\times$21 = 945. The difference between 1134 and 945 is 189, which is a bit smaller than 198. So the number must have been a bit more than 21.

54$\times$22 = 1188 and 45$\times$21 = 990. The difference between 1188 and 990 is 198! So the number must have been 22.

54$\times$10 = 540 and 45$\times$10 = 450. The difference between 540 and 450 is 90, which is much smaller than 198. So the number must have been much greater than 10.

54$\times$20 = 1080 and 45$\times$20 = 900. The difference between 1080 and 900 is 180, which is a bit smaller than 198. So the number must have been a bit more than 20.

Using 20 instead of 10, the difference doubled, so the difference increased by the same proportion as the numbers we used. 20 gave us a difference of 180, which is 18 less than the difference we wanted. 18 is 10% of 180, so to increase the difference by 10%, we should multiply by a number that is 10% more than 20.

10% of 20 is 2, so Jane must have multiplied by 22.

To check, 54$\times$22 = 1188, 45$\times$21 = 990, and the difference between 1188 and 990 is 198!

54 is 45 + 9. So multiplying by 54 is the same as multiplying by 45 and multiplying by 9, and adding the answers together.

So when Jane multiplied by 54 instead of 45, she got 9 extra lots of the number she multiplied by.

So 9$\times\text{the number}$ must be 198, so $\text{the number}$ must have been 198$\div$9 = 22.

Let the number that Jane multiplied by be called $n$. Then $54n$ is $198$ more than $45n$.

So $54n=45n+198$, so $54n-45n=198$, so $9n=198$, so $n=198\div9=22$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.