Number System - Questions & Discussions

MBA CAT 2022
3116
@newavtar said: i have some number.if i divide by 2 ,reminder 1,if i divide it by 3,also remain reminder 1,divide by 4,also reminder 1 ,divide by 5,also reminder 1,divide by 6 ,reminder 1 and finally when divide by 7 reminder 0..what is the number?
is the answer 301???
3117
@Squib said:is the answer 301???
@newavtar said:i have some number.if i divide by 2 ,reminder 1,if i divide it by 3,also remain reminder 1,divide by 4,also reminder 1 ,divide by 5,also reminder 1,divide by 6 ,reminder 1 and finally when divide by 7 reminder 0..what is the number?

The funda is that try 7*x3..put x=1,2,3...coz no. ending with 1 will have rem. 1 when div. by 5....x8 would have done but in that case it would have been div. by 2

3118
@newavtar said:i have some number.if i divide by 2 ,reminder 1,if i divide it by 3,also remain reminder 1,divide by 4,also reminder 1 ,divide by 5,also reminder 1,divide by 6 ,reminder 1 and finally when divide by 7 reminder 0..what is the number?

Many numbers satisfy this criteria until say you ask for the least such number...

Anyway, numbers that leave a remainder of 1 when divided by 2,3,4,5,6 is LCM(2,3,4,5,6)k + 1 which is 60k + 1. For 60k + 1 to be completely divided by 7, the least value of k is 5. Hence the least such number is 60*5 + 1 = 301

3119

CORRECT

3120

Is the answer c i.e only 28

3121

hi..

needed a little help...
to express a number as sum of consecutive nos...
to find the number of ways it can be expressed..
is this approach correct...
find the number of odd factors of the number and then subtract 1 from it...
i.e.
for 21...factors are 1 ,3,7,21
so the number of ways 21 can be expressed as sum of consecutive numbers is 4-1=3

is the approach acceptable?
3122

@[448438:anudsweet]


is the answer 8?
3123

Kindly help, How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers ??Thank you.

@Enceladus said: The answer is nothing but all odd numbers -> 499. and all multiples of 4 -> 249. => Total = 499 + 249 = 748. .
Whats the logic??
3124

No..For the 1st que answer is 3 and 2 que amswer is 6..U can see the solutions posted by others in previous pages

@saurav205 said: @anudsweet is the answer 8?
3125
@mbajamesbond said: Kindly help, How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers ??Thank you. Whats the logic??
all the odd numbers and multiples of four can be represented as the difference of the squares of natural numbers.

so ans = 500 + 250 = 750 should be the answer.
3126

what is the remainder when 50^51^52 is divided by 11?????

3127

@[571364:saiky1991] is it 0?


3128
@saiky1991 said: what is the remainder when 50^51^52 is divided by 11?????

Is it 3?


3129
@saiky1991 said: what is the remainder when 50^51^52 is divided by 11?????
50^51^52 divided by 11:

N ^ 10 mod 11 = 1 (where N is co-prime to 11)

Now, consider 51^52, we need to express it as 10K + r
and as the units digit of 51^52 is 1, so

51^52 = 10K + 1

so, now we find 50 ^ (10K + 1) mod 11

50^10K mod 11 = 1
50^1 mod 11 = 6

So, the remainder is 6.
3130

@[517437:Missioncat2012]..there was mistake in calculation....is it 6?

3131

@[37213:jain_ashu] plz i saw in earlier threads u hv explnd qs lyk 51^203 /7 plz can u expln it agn


3132
@sahib_sheetu said: @jain_ashu plz i saw in earlier threads u hv explnd qs lyk 51^203 /7 plz can u expln it agn
plz any one out dere can help me in this .............
dnt knw much newly joined
3133
@sahib_sheetu said: plz any one out dere can help me in this .............dnt knw much newly joined
Take the same question, 51 leaves 2 as remainder when divided by 7.

So 51^203 will leave 2^203 as remainder when divided by 7.

2^3 leaves 1 as remainder when divided by 7.

so (2^3)^67=2^201 will leave 1 as remainder when divided by 7.

2^2 will leave 4 as remainder when divided by 7.

thus 51^203 will leave a remainder of 4 when divided by 7.
3134

@[595194:sahib_sheetu]... I think you got your answer. I, now having completed my MBA, visit this website infrequently nowadays so my responses maybe delayed.

Enjoy preparing on PG, its the best part. Cheers!

3135

@[592846:YouMadFellow] @[474787:deedeedudu]


50^10=1(mod11)
=>50^50=1(mod11)
=>50^51=6(mod11)
=>50^51^51=6^51(mod11)

6^50=1(mod11)
6^2=3(mod11)
6^52=3(mod11)

The remainder should be 3. How come the answer is 6, which is correct.
Please clear karein.

and isn't 50^51^52 equal to 50^2652

http://www.wolframalpha.com/input/?i=50%5E51%5E51+mod+11

http://www.wolframalpha.com/input/?i=50%5E2652+mod+11

why is there a difference between the two results

Dimaag slow ho gaya hai!! Sona chahiye ab