Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 61)
61.
When a number is divided by 13, the remainder is 11. When the same number is divided by 17, then remainder is 9. What is the number ?
Answer: Option
Explanation:
x = 13p + 11 and x = 17q + 9
13p + 11 = 17q + 9
17q - 13p = 2
| q = |
2 + 13p |
| 17 |
| The least value of p for which q = | 2 + 13p | is a whole number is p = 26 |
| 17 |
x = (13 x 26 + 11)
= (338 + 11)
= 349
Discussion:
47 comments Page 4 of 5.
Prahlad said:
6 years ago
Can u please solve how P=26
Bindu said:
6 years ago
Check from the options,
A) When 339 is divided by 13 we get the remainder as 1 but are should be 11 so check with the next option.
B) When 349 is divided by 13 we get the remainder as 11 and when it is divided by 17 are is 9, it satisfies both the conditions given in the question hence this is the answer.
A) When 339 is divided by 13 we get the remainder as 1 but are should be 11 so check with the next option.
B) When 349 is divided by 13 we get the remainder as 11 and when it is divided by 17 are is 9, it satisfies both the conditions given in the question hence this is the answer.
PNP BIT said:
6 years ago
We can also find;
X=13p+11.
We can check values frm option and put in x.
i.e 349=13p+11.
338=13p.
Then check divisibility of 13.
If it is divisible by13 then it is the right answer.
Because 11 is the remainder if we subtract from the no. We will get an answer which will be exactly divisible by 13.
X=13p+11.
We can check values frm option and put in x.
i.e 349=13p+11.
338=13p.
Then check divisibility of 13.
If it is divisible by13 then it is the right answer.
Because 11 is the remainder if we subtract from the no. We will get an answer which will be exactly divisible by 13.
MUKESH said:
5 years ago
Why can't we take 128 as answer, as it fulfill all the required conditions. But option is not given.
Pragati said:
5 years ago
P=9 also works and answer=128.
Suchismita said:
5 years ago
Please explain, how we get p=26?
Shahbaj said:
5 years ago
26 is a number by which if we use in a equation of q it is completely divisible by 17.
q=2+13(26) /17=20 that's why we take p=26.
Similarly if we put q=20 in eq.
P=17(20) -2/13=26,
So in both way
X=13(26) +1=349.
Y=17(20) +9=349.
q=2+13(26) /17=20 that's why we take p=26.
Similarly if we put q=20 in eq.
P=17(20) -2/13=26,
So in both way
X=13(26) +1=349.
Y=17(20) +9=349.
(2)
Madhu said:
5 years ago
Let X be the number which we have to find.
From given data we make two equations.
13k+11=X ---> (1)
17m+9 =X ---> (2)
From eq (1).
i.e.
13k+11 = X.
Adding 2 on both sides to make 11 as 13.
13k+11+2=X+2,
13k+13=X+2,
13(k+1)=X+2,
k+1=(X+2)/13,
From the options, X=349 is the only option is exactly dividable by 13.
Like vice eq(2).
Add 8 on both sides to 17.
At last, we get:
m+1=(X+8)/17.
So, X=349 is exactly dividable by 17.
From given data we make two equations.
13k+11=X ---> (1)
17m+9 =X ---> (2)
From eq (1).
i.e.
13k+11 = X.
Adding 2 on both sides to make 11 as 13.
13k+11+2=X+2,
13k+13=X+2,
13(k+1)=X+2,
k+1=(X+2)/13,
From the options, X=349 is the only option is exactly dividable by 13.
Like vice eq(2).
Add 8 on both sides to 17.
At last, we get:
m+1=(X+8)/17.
So, X=349 is exactly dividable by 17.
Satya said:
5 years ago
13p + 11 = 17q + 9.
q = (2 + 13p)/17.
To find the least value of p for which q = (2 + 13p)/17 is a whole number; The divisor must completely divide the dividend.
( i.e. denominator must completely divide the numerator.)
So the possible dividends are 170, 340, 510, ....
With Dividend 170,
(2 + 13p) = 170------------> p = 12.9..... , i.e. a fraction.
With dividend 340,
(2 + 13p) = 340 ------------> p = 26 i.e. a whole number.
q = (2 + 13p)/17.
To find the least value of p for which q = (2 + 13p)/17 is a whole number; The divisor must completely divide the dividend.
( i.e. denominator must completely divide the numerator.)
So the possible dividends are 170, 340, 510, ....
With Dividend 170,
(2 + 13p) = 170------------> p = 12.9..... , i.e. a fraction.
With dividend 340,
(2 + 13p) = 340 ------------> p = 26 i.e. a whole number.
(2)
Sudhanshu said:
5 years ago
Here is a simpler way for all of you.
Let no. be N, then;
N = 13a+11. > N-11 =13a (eq. 1).
N = 17b+9. > N-9 =17b (eq. 2).
Thus eq 1 and 2 suggest that N-11 should be divisible by 13 and N-9 should be divisible by 17.
N=349 satisfies both conditions!
Let no. be N, then;
N = 13a+11. > N-11 =13a (eq. 1).
N = 17b+9. > N-9 =17b (eq. 2).
Thus eq 1 and 2 suggest that N-11 should be divisible by 13 and N-9 should be divisible by 17.
N=349 satisfies both conditions!
(14)
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