Latest TCS Aptitude Question SOLUTION: how many 3 digit no which on dividing by 5,6,7 leaves remainder 2,3,5

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(#M40025431) TCS Logical Reasoning question Keep an EYE

how many 3 digit no which on dividing by 5,6,7 leaves remainder 2,3,5

Asked In TCS vertika srivastav (10 years ago)

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veerapuraju pavan ajay  (10 years ago)

for 5 the 3 digit no last digit must 2,7 then only remainder will be 2
for 6 the 3 digit no last digit must 1,3,5,7,9 then only remainder will be 3
for 7 the 3 digit no last digit must 1,2,3,4,5,6,7,8,9 then only remainder will be 2
compare all last digits we have only 7 as common so last digit must be 7
for 7 if we want 5 as remainder so digits must 117,259,327,397,467,537,617,677,747,,827,887,957
from above digits digits which give 2&3 as remainders when number divided by 5&6 are 117,327,537,747,957 so total of 5 digits
answer 5

Is this solution Helpfull? Yes (37) | No (6)  

sakshi sahni  (10 years ago)

LCM of 5,6,7 is 210, so the number will be exactly divisible with difference of 210. similary with remainder 2,3,5 will be on difference of 210
numbers divisible by 5 leaving remainder 2 are: 102,107,112,117.....
number divisible by 6 leaving remainder 3 are 117......
number divisible by 7 leaving remainder 5 are 117.......
so the first three digit no by 5 leaving remainder 2 and 6 leaving remainder 3 and 7 leaving remainder 5 is : 117
the difference is :210
so the nos are : 117,327,537,747,957
hence there are 5 such nos..
ans is 5

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Ishan  (10 years ago)

Numbers that leave remainder 2 on dividing by 5 are:
102,107,112,......997
Quotients on dividing by 5 are
->> 20,21,22,23.....199

now out of (20,21,22....199) numbers which give remainder as 3 on division are
21,27,33,39........195
quotients obtained on dividing these numbers by 6 are:
3,4,5,6,7,8,.....32

now out of (3,4,5,6...32) numbers which give 5 as remainder on division by 7 are:
12,19,26 only

so answer should be 3

please give your valuable feedback .

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surbhi rathaur  (10 years ago)

numbers divisible by 5 leaving remainder 2 are: 102,107,112,117,.......997(differ by 5)
common numbers divisible by 5 nd 6 leaving remainder 2 nd 3 respectively are:117,147,177,207,237,267,297,327.......987.(differ by 30)
now out of these the common numbers divisible by 5,6,7 leaving remainder 2,3,5 respectively are:117,327,537,747,957(differ by 210).
so the answer is 5.

Is this solution Helpfull? Yes (18) | No (1)  

Rohini bhargava  (10 years ago)

general no is of d form:210x-2 put x=1,2,3,4
ans=4

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HARSH  (10 years ago)

YES THERE ARE ONLY 5 OPTION...117,324,537,747,957
LOGIC IS IF U DIVIDE A NUMBER BY 5 AND IT LEAVES A REMAINDER 2 MEANS NUMBER HAVING LAST DIGIT EITHER 7,2...AND IF SAME NUMBER DIVIDED BY 6 LEAVES REMAINDER 3 MEANS MULTIPLE OF 6 HAVING LAST DIGIT EITHER 4,9...9 NEVER COMES IN LAST DIGIT IN MULTIPLE OF 6 SO ALL THE OPTION HAVING ONLY LAST DIGIT 7 GUYS IT'S JUST A HINT TRYING YOURSELF.

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Surabhi  (10 years ago)

n*LCM(5,6,7)-3
put n=1,2,3,4
ans=4

Is this solution Helpfull? Yes (3) | No (8)  

puja  (10 years ago)

117,327,537,747,957 thus 5 in all

Is this solution Helpfull? Yes (3) | No (5)  

Abhishek Bhattacharyya  (9 years ago)

When a three digit number is divided by
5,6,7 successively, the final quotient will be
a single digit (as 5*6*7 = 210 and the max
three digit number 999 is not more than
210*5).
Let the number be x and the quotient after
division by 5 be a and the quotient after
division by 6 be b and the quotient after
division by 7 be c.
Then, x = 5a+2,
a = 6b +3
b = 7c + 5.
Now start with c=1, b=12, a=75, x=377
when c=2, b=19, a=117, x=587,
c=3, b=22, a=113, x=454 and so on with
an increment of 210. Thus the 3 digit
numbers possible are 377,587,797
The answer is 3 such numbers are
possible.

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