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How many three-digit numbers are not divisible by 5, have digits that sum to less than 20, and have the first digit equal to the third digit?
(A) 52
(B) 60
(C) 66
(D) 68
(E) 70
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(A) 52
(B) 60
(C) 66
(D) 68
(E) 70
_________________
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Bunuel
good question; hope not to get such questions on actual day
condition given ; first and last place has to be same
so from 100 to 400 we will have ; 1*10*1 ; 10 such no not divisible by 5, have digits that sum to less than 20, and have the first digit equal to the third digit
so 10 *4 ; 40
all of 500 series is ruled out
for 600 series ; 1*8*1 = 8 such no
for 7 ; 6
for 8 ; 4
for 9 ; 2
tota; 40+8+6+4+2 ; 60
IMO B
General Discussion
Posts: 4,846
Conditions here are that the first digit is equal to the last digit and since the numbers are not divisible by 5, the last digit cannot be 5 (or 0).
The first digit therefore also cannot be 5. 0 is automatically eliminated as 3 digit numbers do not start with 0.
The first place can be any of the 8 digits (1, 2, 3, 4, 6, 7, 8 or 9)
The second place can be any of the 10 digits from 0 to 9.
The last place will be the same as the first place, and therefore the number for this position has been already chosen. The number of ways of filling the last place is = 1.
Total numbers possible = 8 * 10 * 1 = 80
These will also include numbers whose sum of digits is \(\geq\) 20
Numbers starting with 9 - 999, 989, 979, 969, 959, 949, 939, 929 = 8 numbers
Numbers starting with 8 - 898, 888, 878, 868, 858, 848 = 6 numbers
Numbers starting with 7 - 797, 787, 777, 767 = 4 numbers
Numbers starting with 6 - 696, 686 = 2 numbers
Total numbers whose sum is \(\geq\) 20 = 8 + 6 + 4 + 2 = 20
Therefore the total numbers which satisfy the 3 constraints = 80 - 20 = 60
Option B
Arun Kumar
The first digit therefore also cannot be 5. 0 is automatically eliminated as 3 digit numbers do not start with 0.
The first place can be any of the 8 digits (1, 2, 3, 4, 6, 7, 8 or 9)
The second place can be any of the 10 digits from 0 to 9.
The last place will be the same as the first place, and therefore the number for this position has been already chosen. The number of ways of filling the last place is = 1.
Total numbers possible = 8 * 10 * 1 = 80
These will also include numbers whose sum of digits is \(\geq\) 20
Numbers starting with 9 - 999, 989, 979, 969, 959, 949, 939, 929 = 8 numbers
Numbers starting with 8 - 898, 888, 878, 868, 858, 848 = 6 numbers
Numbers starting with 7 - 797, 787, 777, 767 = 4 numbers
Numbers starting with 6 - 696, 686 = 2 numbers
Total numbers whose sum is \(\geq\) 20 = 8 + 6 + 4 + 2 = 20
Therefore the total numbers which satisfy the 3 constraints = 80 - 20 = 60
Option B
Arun Kumar
Posts: 3,656
Given
To Find
Approach and Working Out
Correct Answer: Option B
_________________
- • Three digit numbers with digit sum less than 20.
• First digit is same as the third digit.
To Find
- • The total number of such number.
Approach and Working Out
- • Thee digit numbers that have first digit and the last digit same.
• Let the form be ABA.
- o A = 9, B can be 0 or 1.
o A = 8, B can be 0, 1, 2 or 3.
o A = 7, B can have 6 values.
o A = 6, B can have 8 values.
o A = 5, 4, 3, 2, 1 – B can have 10 values.
• Total such numbers = 2 + 4 + 6 + 8 + 5 × 10 = 70
• Numbers where A is 5 = 10.
- o These are the ones which are multiples of 5.
• Answer = 70 – 10 = 60
Correct Answer: Option B
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