AndrewN wrote:
Bunuel wrote:
How many of the three-digit numbers are divisible by 7?
A. 105
B. 106
C. 127
D. 128
E. 142
Another approach: This one starts with the realization that there are 900 three-digit numbers. After that, mental math can get us the answer quickly and cleanly. It may not be evident how many times 7 goes into 900 evenly, but we should
know that 7 goes into 700 exactly 100 times. With the bulk of the 900
numbers out of the way, we just have to figure out how many times 7 goes into the remaining 200 numbers evenly. Rather than labor over the exact answer, we can see that 7 would fit into 210
exactly 30 times. But 700 (100 sevens) + 210 (30 sevens) = 910 (130 sevens), and we have no more than 900 numbers to work with instead. From 910, just subtract one 7 at a time until the number falls at or below 900: 910 - 7 = 903; 903 - 7 = 896. Since we had 130 sevens that fit into 910 numbers and needed to take away two sevens to give us a valid number, we know that 130 - 2 sevens, or 128 sevens, will be our answer.
(D) it is, then. To recap:
* There are 900 three-digit numbers.
\(700/7=100\) (sevens)
\(210/7=30\) (sevens)
\(700+210=910\) (100 + 30 = 130 sevens)
\(910-2(7)=896\) (130 - 2 = 128 sevens)
This problem (or one similar to it, asking about a different divisor) can be broken down easily in under a minute, and nothing beats
knowing your answer is correct.
Good luck with your studies.
- Andrew
I completely concur with your method
AndrewN, it is a foolproof method. Folks, if you understand this method, then this method will also help you to gain confidence in the Number Properties topic.
I followed the same reasoning, with a slightly different approach as follows -
Last 2 digit number which is divisible by 7 = 98 (14 x 7)
First 3 digit number which is divisible by 7 = 105 (15 x 7) OR (98 + 7)
To find the last 3 digit number without multiple trials and errors, we know that it will be some number between 990 - 999
So, by splitting a value greater than 900 and between 990 - 999 that is divisible by 7, we get -
700 + 98 + 98 + 98 = 994 (Last number)
We can then solve the remaining part using
Bunuel method OR we can reason as follows-
The last 3 digit number which is divisible by 7 = 994,
So 994/7 = 142. This means that the number 7 is added 142 times.
This also means that the first number in this series will be the number 7, which is a one-digit number as well as, other 2-digit numbers which are multiples of 7. But the question stem specifically asks for 3-digit numbers.
So, excluding all 1-digit and 2-digit numbers i.e numbers less than 100. There are 14 such numbers (14 x7 = 98).
Hence, 142 (Multiples of 7 less than 1000) - 14 (Multiples of 7 less than 100) = 128.
This reasoning looks difficult to comprehend at first, but it will definitely help us in getting better with our intuition while solving number property questions.
I hope it helps!