Of three digit positive integers whose three digits are all different and non zero, how many are odd integers greater than 700
1)84
2)91
3)100
4)105
5)243
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Neilsheth2
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Hi Neilsheth2,
You'll likely find it easiest to handle this calculation in 'pieces'
This question lays out the following restrictions:
1) 3-digit numbers greater than 700
2) All digits are NON-0 and DIFFERENT
3) The number must be ODD.
Let's start with the 700s...
1st digit is 7 = 1 option
3rd digit must be ODD, but NOT 7 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 700s
Now, the 800s...
1st digit is 8 = 1 option
3rd digit must be ODD = 5 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(5)(7) = 35 numbers in the 800s
Finally, the 900s; the math here works the same as the 700s...
1st digit is 9 = 1 option
3rd digit must be ODD, but NOT 9 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 900s
28+35+28 = 91
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
You'll likely find it easiest to handle this calculation in 'pieces'
This question lays out the following restrictions:
1) 3-digit numbers greater than 700
2) All digits are NON-0 and DIFFERENT
3) The number must be ODD.
Let's start with the 700s...
1st digit is 7 = 1 option
3rd digit must be ODD, but NOT 7 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 700s
Now, the 800s...
1st digit is 8 = 1 option
3rd digit must be ODD = 5 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(5)(7) = 35 numbers in the 800s
Finally, the 900s; the math here works the same as the 700s...
1st digit is 9 = 1 option
3rd digit must be ODD, but NOT 9 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 900s
28+35+28 = 91
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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Neilsheth2
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GMATGuruNY
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Case 1: Hundreds digit is 7 or 9Neilsheth2 wrote:Of three digit positive integers whose three digits are all different and non zero, how many are odd integers greater than 700
1)84
2)91
3)100
4)105
5)243
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (Any odd digit but the one in the hundreds place)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
2*4*7 = 56.
Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 5. (Any of the 5 odd digits)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
1*5*7 = 35.
Total options = 56+35 = 91.
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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