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its always better to go for a licenesed antivirus , you can pass a year without the tension of any viruses , kaspersky is a great antivirus, i buy it every year, using it from last 3 years, no problem ever ::

its always better to go for a licenesed antivirus , you can pass a year without the tension of any viruses , kaspersky is a great antivirus, i buy it every year, using it from last 3 years, no problem ever

please solve this

what is x if

what is x if

> It can have multiple answers
it is quite clear that
S = 9
m = 1
R = 8
O=0
e+1 = n
now e can be 3/4/5;n ==> 4/5/6
thankyou brother:D

It can have multiple answers

it is quite clear that

S = 9

m = 1

R = 8

O=0

e+1 = n

now e can be 3/4/5;n ==> 4/5/6

thankyou brother:D

Each alphabet stands for a digit. Try and determine which letter stands for which digit.
Please explain also

Please explain also

Can some1 please solve this with approach
Q : What is the least square number which is divisble by 3,5,6 and 9 ? ::

Can some1 please solve this with approach

Q : What is the least square number which is divisble by 3,5,6 and 9 ?

Q : What is the least square number which is divisble by 3,5,6 and 9 ?

> 9456
1087
\------
10543
check your answer , answer almost there
and please post the approach also ::

9456

1087

------

10543

check your answer , answer almost there

and please post the approach also

> my take 4
2*2.....(203 times) %7
(2*2*2) % 7 is 1
so 2*2*2........up to 201 digit remainder is 1
2*2% 7 = 4
correct ::
thankyou brother

my take 4

2*2.....(203 times) %7

(2*2*2) % 7 is 1

so 2*2*2........up to 201 digit remainder is 1

2*2% 7 = 4

correct

thankyou brother

Please explain the method behind the following question ::
Find the remainder when (51)^203 is divided by 7
a) 4
b) 2
c) 1
d) 6

Please explain the method behind the following question

Find the remainder when (51)^203 is divided by 7

a) 4

b) 2

c) 1

d) 6

Find the remainder when (51)^203 is divided by 7

a) 4

b) 2

c) 1

d) 6

> UtsavGambhir Says
>
> It would be 31 and "y" in that case would have 9 cards. If "y" gives "x" 1 card then the count with both would be 8 and 32. When "x" gives 1 card to "y", the count would be 10 and 30.
explanation to this would be highly appreciated ::

UtsavGambhir SaysIt would be 31 and "y" in that case would have 9 cards. If "y" gives "x" 1 card then the count with both would be 8 and 32. When "x" gives 1 card to "y", the count would be 10 and 30.

explanation to this would be highly appreciated

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