Latest CAT Aptitude Question SOLUTION: How many 5-digit numbers are there such that the digits at hundred place, unit place and ten-thousand place form a gp? ans =3300 how?

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How many 5-digit numbers are there such that the digits at hundred place, unit place and ten-thousand place form a gp? ans =3300 how?

Asked In CAT raktim (9 years ago)

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ANSHUL KUMAR  (9 years ago)

[Swarnabha Dey]
if the digits of unit places,hundred places and ten places have to form a gp, Then it should consist of 111,124,421,139,931,248,842,222,333,444,555,666,777,888,999,000.
now its given that the number is a five f=digit number so assuming 111, the 1st two digits can be filled in 9*10 = 90 ways,
so 16 possible combinations are there ,total number of combinations will be = (16*90) =1440

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Swarnabha Dey  (9 years ago)

if the digits of unit places,hundred places and ten places have to form a gp, Then it should consist of 111,124,421,139,931,248,842 .
now its given that the number is a five f=digit number so assuming 111, the 1st two digits can be filled in 9*10 = 90 ways,
so 7 possible combinations are there ,total number of combinations will be = (7*90) =630

Is this solution Helpfull? Yes | No (2)  

Anurag Yadav  (9 years ago)

73300
Let hundred unit and ten thousand no be a,ar,ar2 then
If a=1 then r has 2 possible value 2,3 and no formed=400
If r=1 then a cn hv 9 values so nos possible=72900
Total =73300

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Abhishek Mohanty  (7 years ago)

let the number be _ _ _ _ _

if the number at ten-thousands place is:-

1 then the other digits can be (1,1),(2,4),(3,9),(4,2),(9,3)----5

2............(2,2),(4,8),(8,4)----------------------------------3

3............(3,3)..............................................1

4............(4,4),(2,1),(1,2)..................................3

5............(5,5)..............................................1

6............(6,6)..............................................1

7............(7,7)..............................................1

8............(8,8),(4,2),(2,4)..................................3

9.............(9,9),(3,1),(1,3).................................3

hence total no of distinct numbers are 21

now for each of these 21 numbers there are 10 digits at tens place

and 10 digits at thousands place so total answer is 21*10*10=2100(ans)

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lakhjeet singh  (5 years ago)

abcde be the five digit number,
for terms to be in gp ,middle term^2 = product of other two terms
(1,2,4) will give 6 solutions
(1,3,9)-6 solutions
(2,4,8)-6 solutions
(4,6,8)- solutions
these 6 cases are for 0nly 3 terms that are in GP for rest of teh two places any of the 10 possible numbers can fit
so total possible combinations = (6*4*100) = 2400
now (1,1,1)(2,2,2)....(9,9,9) another 900 cases
total combination = 2400+900 =3300

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