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01 Dec 2021, 06:01
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On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of \($a6,79b\) where \(a\) is the ten thousands digit and \(b\) is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of \(a + b\) ?
A. \(4\)
B. \(5\)
C. \(6\)
D. \(9\)
E. \(10\)
A. \(4\)
B. \(5\)
C. \(6\)
D. \(9\)
E. \(10\)
01 Dec 2021, 06:01
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Official Solution:
On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of \($a6,79b\) where \(a\) is the ten thousands digit and \(b\) is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of \(a + b\) ?
A. \(4\)
B. \(5\)
C. \(6\)
D. \(9\)
E. \(10\)
To solve this question, we should know divisibility rule for 8 and 9.
Divisibility rule for \(2^n\):
An integer is divisible by 2 if its last digit is divisible by 2.
An integer is divisible by 4 if its last two digits is divisible by 4.
An integer is divisible by 8 if its last three digits is divisible by 8.
...
Generally, an integer is divisible by \(2^n\) if its last \(n\) digits is divisible by \(2^n\). ...
Divisibility rule for \(9\):
The rule for divisibility by 9 is similar to divisibility rule for 3. That is, if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.
Back to the question:
We are given that \(72*p=a6,79b\), where \(p\) is an integer price of each figure.
\(p=\frac{a6,79b}{8*9}=integer\). So, \(a6,79b\) must be divisible by both 8 and 9.
According to the rule above, \(a6,79b\) will be divisible by 8, whenever \(79b\) is divisible by 8. 800 is divisible by 8, next smallest number divisible by 8 is 792, so \(b=2\)
\(a6,79b\) will be divisible by 9, if \(a+6+7+9+b=a+6+7+9+2=24+a\) is divisible by 9, so when \(a=3\).
Hence, \(a+b=2+3=5\).
Answer: B
On a San Diego Comic-Con, Sheldon bought all 72 identical collectible Mr. Spock action figures. He paid a total of \($a6,79b\) where \(a\) is the ten thousands digit and \(b\) is the units digit. If each figure cost the same and the cost in dollars was an integer, what is the value of \(a + b\) ?
A. \(4\)
B. \(5\)
C. \(6\)
D. \(9\)
E. \(10\)
To solve this question, we should know divisibility rule for 8 and 9.
Divisibility rule for \(2^n\):
An integer is divisible by 2 if its last digit is divisible by 2.
An integer is divisible by 4 if its last two digits is divisible by 4.
An integer is divisible by 8 if its last three digits is divisible by 8.
...
Generally, an integer is divisible by \(2^n\) if its last \(n\) digits is divisible by \(2^n\). ...
Divisibility rule for \(9\):
The rule for divisibility by 9 is similar to divisibility rule for 3. That is, if the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.
Back to the question:
We are given that \(72*p=a6,79b\), where \(p\) is an integer price of each figure.
\(p=\frac{a6,79b}{8*9}=integer\). So, \(a6,79b\) must be divisible by both 8 and 9.
According to the rule above, \(a6,79b\) will be divisible by 8, whenever \(79b\) is divisible by 8. 800 is divisible by 8, next smallest number divisible by 8 is 792, so \(b=2\)
\(a6,79b\) will be divisible by 9, if \(a+6+7+9+b=a+6+7+9+2=24+a\) is divisible by 9, so when \(a=3\).
Hence, \(a+b=2+3=5\).
Answer: B
24 Jul 2023, 23:39
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After looking at the options - just the divisibility check for 9 is enough.
a6,79b
6+7+9 = 22. => 27 is the next multiple of 9.
So you need either a + b = 5 or a + b = 5 + 9 = 14.
Since 14 is not there, a+b has to be 5.
a6,79b
6+7+9 = 22. => 27 is the next multiple of 9.
So you need either a + b = 5 or a + b = 5 + 9 = 14.
Since 14 is not there, a+b has to be 5.
