How many multiples of 7 between 100 and 150?

Question

Hey everyone. As you can see, I need help with something so simple. I'm sure the answer is right in front of me and I can't see it. They use the above question in Manhattan Number Properties. They start listing multiples: 105, 112,119,126, 133, 140, 147. The thing that is not clear to me is how I know to start with 105 instead of say, 103 or 101.

dabral · Accepted Answer

Response in attached images.

bb · Answer

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105). 
The one before that is 98, and that's less than 100.

This is the most reliable way to solve the question (listing). 
Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Hope this helps.

gmatz101 · Answer

Wow. Makes perfect sense. Thanks a lot.

Tgt750 · Answer

My thought process:
Smallest multiple > 100 = 105, multiple 105/7 = 15
greatest multiple < 150 = 147, multiple 147/7 = 21, Therefore total number of multiples = 21-15 +1 = 7

dabral · Answer

Slightly different approach, once the beginning and ends are established.

bb · Answer

Hm.... so that would be 
Step 1: (150 - 100) = 50
Step 2:  \frac{50}{7}  = 7
Step 3: 7+1 = 8 ?

I think you meant (Last - First +1)/Multiple

akankshabms · Answer

Its well explained in Manhattan word problems book

Posted from my mobile device

onewayonly · Answer

Why does the (last - first) / increment rule not always work here? I know it doesn't (for example, count the number of multiples of 7 between 11 and 24. There are two (14, 21), but (last - first) / increment gives you 1.

Also, the Manhattan GMAT book lists (last - first) / increment + 1 as the rule to use here, but that doesn't always work either. For example, count the number of multiples between 9 and 24. There are two: 14, and 21. But (last - first) / increment + 1 is gives you 3 multiples.

It appears that in these sorts of problems involving the number of multiples between two numbers: we cannot strictly rely on formulas and instead need to think them through on a case by case basis, right?

Thanks.

Im2bz2p345 · Answer

The formula is NOT just "last - first," but the actual formula is "last MULTIPLE in range - first MULTIPLE in range." Here is the complete formula: # \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1 . You can refer to Bunuel's explanation as he uses it in multiple examples here: how-many-multiples-of-4-are-there-between-12-and-94862.html#p730075 Using one of your examples, you stated: "count the number of multiples of 7 between 11 and 24. There are two (14, 21), but (last - first) / increment gives you 1." This would be solved like this: \frac{21 - 14}{7}+1 = 2 Using your other example, you stated: "count the number of multiples between 9 and 24. There are two: 14, and 21. But (last - first) / increment + 1 is gives you 3 multiples." I'm assuming you meant "count the number of multiples of 7" between the range of 9 and 24 (you didn't specify the multiple in this example). This would be solved using the exact same numbers as shown in the above example. Once again, the answer would be 2. Hope this helps, ~ Im2bz2p345

dabral · Answer

@onewayonly 
The last and first term refer to the terms of an arithmetic sequence, and you have to make sure that all the terms are part of an arithmetic sequence, where the difference between successive terms is fixed. In the case of counting multiples, you have to start with the first multiple in the range and the last multiple in the range, we can't just use the first and last term of the given range.

Cheers,
Dabral

samirchaudhary · Answer

For such questions : Last term is the last multiple within the range and the first term is the first multiple within the range.
This series of multiples forms an AP sequence.

Lets understand the formula  ((Last term - First term)/ spacing) +1 
We get this formula from AP(Arithmetic Progression) formula for last term i.e Tn = a + (n - 1) d

Lets take the first term as 'a' and the nth term as 'b'

Using the formula for nth term of an AP, we get : b = a + (n - 1)d. We need the value of 'n'.
Solving we get :  n = ((b - a)/d)+1 
where,
b = last term , a = first term, and d = common difference.

I hope this helps.

Nunuboy1994 · Answer

OK so we can just manipulate the formula

(largest multiple of 7 between 100 and 150 - smallest multiple of 7 between 100 and 150)/ 7 + 1 = 7

This is derived from the formula for the number of even numbers in a consecutive  sequence

largest even number- smallest even number / 2 +1

bellabear · Answer

bb  
Could you please explain where this method is from? and why it does not work very well? Thank you! >> Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Madavco · Answer

So in the question, the multiples of 7 between 100 and 150 are: 105, 112, 119, 126, 133, 140, and 147 which means that there are seven multiples of 7 between 100 and 150

(150-100)/7 = 50/7 = 7 remainder 1, so if you drop the remainder you would get the same answer as we did counting above by subtracting the upper limit of the range from the lower limit and dividing by the factor.

bb says this is a dangerous method, because what if I asked you "how many multiples of 7 are there between 104 and 155?" The answer would be 105, 112, 119, 126, 133, 140, 147, and 154 which equals eight.

Using the alternative method, I would get (155 -104)/7 = 51/7 = 7 remainder 2, dropping the remainder gives me an answer of seven which is the wrong answer to my new question.

I hope this adds some clarity to why bb said this method does not work very well, or at least it does not work very reliably. In the original case you would get the correct answer, however in the similar case I presented you would get the wrong answer.

JaeTPU · Answer

The most efficient way would be to recognize that 105=70+35. You can use the following approach if it doesn't come to you right away:

Divide 100 by 7, your remainder is 2. Since 2+5 is 7, adding 5 to 100 will make it divisible by 7.

AvinashHaran · Answer

If we divide 100 by 7, we get remainder = 2. Since 2+5 = 7, adding 5 to 100 will make it divisible by 7 and that is 105. 
Similarly if we divide 150 by 7, we get remainder = 3, so if we remove 3 from 150 we get the largest number less than 150 that is divisible by 150.
To calculate the number of terms in Arithmetic Progression, we use: a+(n-1)d=last term
where a=first term
n=number of terms
d=constant difference between any two successive terms
=>105+(n-1)7=147
=>(n-1)7=42
=>n-1=42/7=6
=>n=7
Therefore the number of multiples of 7 between 105 and 147 (inclusive)= 7

In this example as the number of terms are 7, we can do it manually. However, for a larger sequence we need to use Arithmetic Progression formula.

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How many multiples of 7 between 100 and 150?

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How many multiples of 7 between 100 and 150?

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