@sonamaries7 said:The sum of the cubes of any 3 consecutive nos is always divisible by:18279none
option c..
thanks for asking...
thanks for asking...
@sonamaries7 said:The sum of the cubes of any 3 consecutive nos is always divisible by:18279none
Any no or natural numbers??
If natural numbers, then it should be 9...otherwise none
@prateek987 said:option c..thanks for asking...
@ScareCrow28 said:Any no or natural numbers??If natural numbers, then it should be 9...otherwise none
@Pravs_Destiny said:@sonamaries79???
share ur approach also..
@audiq7 nahi bhai...galat hai. chalo when u feel like doing ..solve kar ke tag kar denaanswer 1036 hai
@sonamaries7 said:share ur approach also..
(k-1)^3 + k^3 + (k+1)^3 = 3(k^3+2k)
Take k^3+2k
If k=3a
It is divisible by 3
k=3a+1
Again, divisible by 3
k=3a+2
Divisible by 3
Hence The product is divisible by 9. (It should be natural numbers)..
find the maximum number of arithmetic progressions of integral terms that can be formed with at most 11 terms and at least 3 terms where first term =37224 and last term =39624??
@pavimai said:find the maximum number of arithmetic progressions of integral terms that can be formed with at most 11 terms and at least 3 terms where first term =37224 and last term =39624??
6?
@sonamaries7
I took any three natural number cubed it and checked 18 and 27 got eliminated left with 9 it was divisible every time..
Took (123) then (234) and (345) since we are aware of their cubes..
I took any three natural number cubed it and checked 18 and 27 got eliminated left with 9 it was divisible every time..
Took (123) then (234) and (345) since we are aware of their cubes..
@pavimai said:share ur approach bhai.its correct
39624 - 37224 = 2400
From 3 to 11 There are 6 factors of 2400 which are 3, 4, 5, 6, 8 and 10
Hence The series can be written in 6 diff forms
@pavimai said:find the maximum number of arithmetic progressions of integral terms that can be formed with at most 11 terms and at least 3 terms where first term =37224 and last term =39624??
37224+(n-1)d=39624
(n-1)d=2400
Nd=2400
N can vary from 2 to 8
6?
except N=7
all others can be made
hence 6 in total?
@pavimai said:find the maximum number of arithmetic progressions of integral terms that can be formed with at most 11 terms and at least 3 terms where first term =37224 and last term =39624??
6 ?
@jashholmes said:In a class, two students weighing 54 and 62 kgs are replaced by two new students weighing 58 and 64 kgs. As a result the average weight of the class increases by one fourth of kg. How many students are there in the class???
24..???
@prateek987 said:@ScareCrow28 what is your secret to total quant dominance..?
I'm just a novice here bhai. But I can tell you one thing, it's all because of PG Quant Thread. You get to learn from others and I did.
@ScareCrow28 said:I'm just a novice here bhai. But I can tell you one thing, it's all because of PG Quant Thread. You get to learn from others and I did.
You are a novice??? Modesty bhai..
Sorry for spamming..
@ScareCrow28 Sir..
I want to know some thing...I have a book called m tyra quicker mathematics ...its this book gud...? it has a lot of shortcuts....
From which do you practice ...?
I want to know some thing...I have a book called m tyra quicker mathematics ...its this book gud...? it has a lot of shortcuts....
From which do you practice ...?
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