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17 Dec 2019, 01:26
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (01:54) correct
30%
(01:53)
wrong
based on 365
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If n is an integer greater than 2 and \(t = 10^n − 132\), if t is represented in decimal notation, which can be the sum of all digits of t?
A. 30
B. 31
C. 32
D. 33
E. 34
Are You Up For the Challenge: 700 Level Questions
A. 30
B. 31
C. 32
D. 33
E. 34
Are You Up For the Challenge: 700 Level Questions
05 Sep 2022, 04:06
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Bunuel
\(10^n\), where n is a non-negative integer, has \(n+1\) digits: 1, followed by \(n\) zeros. For example:
- \(10^0=1\). 1 has one digit: 1, followed by 0 zeros;
\(10^1=10\). 10 has two digits: 1, followed by 1 zeros;
\(10^2=100\). 100 has four digits: 1, followed by 2 zeros;
\(10^3=1,000\). 1,000 has four digits: 1, followed by 3 zeros;
...
\(10^n-132\) will have \(n\) digits: \(n-3\) 9's, followed by 868. For example:
- If n = 3, then \(10^n − 132=868\). The sum of the digits = 22;
If n = 4, then \(10^n − 132=9,868\). The sum of the digits = 31;
If n = 5, then \(10^n − 132=99,868\). The sum of the digits = 40;
If n = 6, then \(10^n − 132=999,868\). The sum of the digits = 49;
...
As we can see, the answer is 31, B.
Answer: B.
If you can deduce the above right away, then you could get the formula for the sum of the digits of \(10^n − 132\) and solve from there. \(10^n − 132\) will have \(n-3\) 9's, followed by 868:
- \(9*(n - 3) + 8 + 6 + 8 = 9n -5\).
So, the answer must be 5 less than a multiple of 9. Only B works: 31 = 4*9 - 5.
Hope it's clear.
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Hope it helps.
General Discussion
17 Dec 2019, 02:15
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Method 1-
\(10^n\)= 1 mod 9, for n>2
132= 6 mod 9
\(10^n − 132\)= (9+1-6) mod 9= 4 mod 9
Only 31 among options gives 4 remainder when divided by 9
Method 2
t=\(10^n-132\)= 999....(n-3) times....868
Sum of digits
= 9+9+9..(n-3)times+8+6+8
= 9(n-3)+22
when n=4, sum of digits = 31
when n=5, sum of digits= 40
B
\(10^n\)= 1 mod 9, for n>2
132= 6 mod 9
\(10^n − 132\)= (9+1-6) mod 9= 4 mod 9
Only 31 among options gives 4 remainder when divided by 9
Method 2
t=\(10^n-132\)= 999....(n-3) times....868
Sum of digits
= 9+9+9..(n-3)times+8+6+8
= 9(n-3)+22
when n=4, sum of digits = 31
when n=5, sum of digits= 40
B
Bunuel
