Problem Solving

How many integers divisible by 3 are there between 10! and 10! + 20 inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10


OA B

number of inetgers between 10! and 10! + 20 inclusive
=10! + 20 -10!+1 (there are a-b+1 integers between integers a and b inclusive)
=21

every 3rd integer is divisible by 3, so there are 21/3=7 integers divisible by 3

the question actually asks how many integers are there divisible by 3 among 21 consecutive integers which starts from an integer whose unit digit is 1.
Take any such 21 consecutive integers and there will be an AP with CD as 3 therefore there can be 7 integers in between

scoobydooby wrote:number of inetgers between 10! and 10! + 20 inclusive
=10! + 20 -10!+1 (there are a-b+1 integers between integers a and b inclusive)
=21

every 3rd integer is divisible by 3, so there are 21/3=7 integers divisible by 3

Hi Scoobydooby could you pls. explain how to count the numbers of integers between those factorials? How can I apply the formula (max+min/2) * (max-min) +1? Tksvm

the number of integers between two integers a and b inclusive is given by the formula: range+1 or (b-a)+1

the numbers of integers between two integers a and b not including a and b is just the range-1

a factorial is an integer. a factorial+integer is an integer

so if a: 10!, b: 10!+20
number of integers between 10!+20 and 10! is therefore (10! + 20 -10!)+1=21

like wise there are 21 integers between 3! and 3!+20 inclusive. (between 6 and 26 inclusive)

but El Cucu, am not aware of the formula you mention: (max+min/2) * (max-min) +1.

scoobydooby wrote:the number of integers between two integers a and b inclusive is given by the formula: range+1 or (b-a)+1

the numbers of integers between two integers a and b not including a and b is just the range: b-a

a factorial is an integer. a factorial+integer is an integer

so if a: 10!, b: 10!+20
number of integers between 10!+20 and 10! is therefore (10! + 20 -10!)+1=21

like wise there are 21 integers between 3! and 3!+20 inclusive. (between 6 and 26 inclusive)

but El Cucu, am not aware of the formula you mention: (max+min/2) * (max-min) +1.

Tks Scooby, I mistook number of integers for sum of numbers ( fhe forumla above is related to the sum of numbers) Sorry for the confusion.

Key point to know is
n! is divisible for all numbers <= n.


So 10! is divisible by 3. ------- (1)

Now forget the factorials and just look at 20.
20/3 = 6 ie. there are 6 numbers that are divisible by 3.
6 + 1 (for the 10! divisible by 3 mentioned in 1)
= 7.

( here 1 is added to 6 coz the problem mentioned 10! is included.
Else the answer is 6)

HT Helps.

scoobydoo

the numbers of integers between two integers a and b not including a and b is just the range: b-a

Are you sure for this, or you wrote that in a hurry. I can't believe that you mean there are 2 integers between 3 and 5, not inclusive!

thanks sanju09 for pointing out, i should have proof read what i wrote.

the numbers of integers between two integers a and b not including a and b is just range-1

sanju09 wrote:How many integers divisible by 3 are there between 10! and 10! + 20 inclusive?

A. 6
B. 7
C. 8
D. 9
E. 10


OA B

Since 10! includes a multiplication of 3 10! must be divisible by 3 so that's one. No 20/3 = 6.

6+1 = 7.

B.

Hi,

A quick warning: be careful when making assumptions that dividing by N will necessarily give you the right number of multiples of N.

If we are talking about 20 consecutive integers, it is possible that the list will contain 6 multiples of 3:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
The multiples of 3 are 3, 6, 9, 12, 15, 18

It is ALSO possible that a list of 20 consecutive integers will contain 7 multiples of 3:

3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22
The multiples of 3 are 3, 6, 9, 12, 15, 18, 21

Instead, make sure that you take note of the SMALLEST and LARGEST multiple of N in the list. This is the key to finding the answer.

In the first example above, the smallest and largest multiples of 3 are: 3 and 18. In the second example above, the smallest and largest multiples of 3 are: 3 and 21.

To find the number of multiples of N in a certain range, first find the SMALLEST and LARGEST multiple of N in the list. Call these A and B.

The number of multiples of N is (B - A)/N + 1.

In the first example above, (B - A)/N + 1 = (18 - 3)/3 + 1 = 5 + 1 = 6.
In the second example above, (B - A)/N + 1 = (21 - 3)/3 + 1 = 6 + 1 = 7.

As another example, think about the multiples of 5 between 6 and 19, inclusive. Then, think about the multiples of 5 between 9 and 16, inclusive. In each case, the answer is 2 (10, 15).

If we just used the differences of the first and last number, we would get:
(19-6)/5 = 2.6
(16-9)/5 = 1.4
This is misleading, since the answer in each case is 2.

If instead we use the formula above, in each case we will find that A = 10, B = 15, and (B - A)/N + 1 = (15 - 10)/5 + 1 = 1 + 1 = 2, which is correct.

Since the question in the initial post asks about 21 consecutive integers, we get a bit lucky, in that any string of 21 consecutive integers has seven multiples of 3.

vittalgmat wrote:Key point to know is
n! is divisible for all numbers <= n.


So 10! is divisible by 3. ------- (1)

Now forget the factorials and just look at 20.
20/3 = 6 ie. there are 6 numbers that are divisible by 3.
6 + 1 (for the 10! divisible by 3 mentioned in 1)
= 7.

( here 1 is added to 6 coz the problem mentioned 10! is included.
Else the answer is 6)

HT Helps.

Great explanation. When I see 10!, I just think of it as this huge random number. It's important to remember that it has specific properties that we can use to solve the question (like divisible by 1-10, even, ends with 2 zeros...)