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For how many of the integers from 10 to 99 is at least one of the two

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Bunuel
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For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  22 May 2020, 07:37
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For how many of the integers from 10 to 99 is at least one of the two digits a 4 ?

A. 9
B. 10
C. 18
D. 19
E. 20


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BrentGMATPrepNow
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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  22 May 2020, 07:45
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Bunuel wrote:
For how many of the integers from 10 to 99 is at least one of the two digits a 4 ?

A. 9
B. 10
C. 18
D. 19
E. 20
PS21197


Number of 2-digit integers with at least one 4 = (total number of 2-digit integers) - (number of 2-digit integers that have ZERO 4's)

total number of 2-digit integers
We want to count the number of integers from 10 to 99 inclusive
A nice rule says: the number of integers from x to y inclusive equals y - x + 1
So the number of integers from 10 to 99 inclusive = 99 - 10 + 1 = 90

number of 2-digit integers that have ZERO 4's
Let's build some 2-digit integers that have ZERO 4's
The tens digit can be 1, 2, 3, 5, 6, 7, 8, or 9 (8 options)
The units digit can be 0, 1, 2, 3, 5, 6, 7, 8, or 9 (9 options)
So, the total number of 2-digit integers that have ZERO 4's = (8)(9) = 72

We get:
Number of 2-digit integers with at least one 4 = 90 - 72 = 18

Answer: C

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Brent
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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  22 May 2020, 07:46
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The integers between 10 to 99,
With at least 1 of the two digits a four.

The no. With 4 as units digit :8 no. ------(1)(14,24,34,54,64,74,84,94)

The no. With 4 as tens digit : 9 no. -------(2)
(40,41,42,43,45,46,47,48,49)

The no. With 4 as both ones and tens digit = 1 -----(3)
(44)

Total no. : 1+8+9=18
Answer is C

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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  22 May 2020, 09:36
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Bunuel wrote:
For how many of the integers from 10 to 99 is at least one of the two digits a 4 ?

A. 9
B. 10
C. 18
D. 19
E. 20


PS21197


FACT: 1 to 99 there are 19 numbers which include digit 4 and 20 times digit 4 is used

But we need to exclude one 4 that comes between 1 and 10

so total numbers with digit 4 used = 19-1 = 18

Answer: Option C
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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  22 May 2020, 09:44
total digits from 10 to 99 ; 99-10 +1 ; 90
and not with 4 we have ( 8*9= 72 terms )
so with 4 we have ; 90-72 ; 18 terms
OPTION C


Bunuel wrote:
For how many of the integers from 10 to 99 is at least one of the two digits a 4 ?

A. 9
B. 10
C. 18
D. 19
E. 20


PS21197
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EgmatQuantExpert
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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  30 May 2020, 22:43
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Solution



To find
We need to determine
    • For how many of the integers from 10 to 99 is at least one of the two digits a 4

Approach and Working out
    • Number of integers in which units digit is 4 = 9 {units digit can take any value from 1 to 9}
    • Number of integers in which tens digit is 4 = 10 {units digit can take any value from 1 to 9}
    • Number of integers in which both the digits are 4 = 1 {44}
    • Thus, total number of such integers = 9 + 10 – 1 = 18

Thus, option C is the correct answer.

Correct Answer: Option C
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Re: For how many of the integers from 10 to 99 is at least one of the two [#permalink] New post  15 May 2022, 23:51
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