Elitmus
Exam
Numerical Ability
Number System
. No when divided by 5, 2, 3 successively in order to get remainder 0, 1, 2.what is remainder
of same no when divided by 2, 3, 5 successively?
Read Solution (Total 22)
-
- 5|X
2|y - 0
3|z - 1
|1 - 2
from here we get, z= 3*1+2=5
y= 2z+1 = 2*5+1=11 and x= 5y+0 = 55
so the number is 55....when it will be divided by 2,3 and 5 the remainders will be 1,0,4 respectively - 10 years agoHelpfull: Yes(45) No(16)
- 3+2 = 5, 5*2 + 1 = 11, 11*5 = 55
smallest no. is 55
when 55 is divided by 2,3,5 we get rem as 1,1,0 - 10 years agoHelpfull: Yes(23) No(45)
- number which successively divided by 5,2,3 and give reminder 0,1,2 is 175
and when the same no divided by 2,3 and 5 it gives reminder 1,0,4 - 10 years agoHelpfull: Yes(22) No(11)
- 5 2 3
0 1 2
now,
2*2+1=5
5*5+0=25
Now,
25/2=qutoint=12 and remainder=1
12/3= qutoint= 4 and remainder=0
4/5 = qutoint=0 and remainder= 4
hence answer is 1 0 4
- 10 years agoHelpfull: Yes(13) No(2)
- ans will be 1,2,0.....
like if we take 35 n then apply given condition
- 10 years agoHelpfull: Yes(12) No(6)
- Let x is the number
x= 5*k+0 (k is qutoint)
k=2*L+1 (L is qutoint)
L=3*M+2 (M is qutoint)
=> K= 2*(3*M+2)+1 = 6M+5
=> x= 5*( 6M+5)+0
=> x =30*M+25;
Put M=1, x=55; M=2, x=85;...so on
let no. is 55 Then,
55/2 = rem= 1 and qusnt=27; 27/3= rem=0 and qusnt=9; 9/5= rem=4 qusnt=1
rem after successively divide by 2, 3, 5 is 1, 0, 4 - 9 years agoHelpfull: Yes(12) No(0)
- As the question info is inadequate
I am taking the number as 5,
when 5 is divided by 5,2,3 it will give a remainder of o,1,2 respectively.
Again when 5 is divided by 2,3,5 it gives a remainder of 1,2,0.
- 10 years agoHelpfull: Yes(7) No(1)
- 85 is the number
85/5 gives 17 as questient and rem=0
now divide 17/2 gives q=8 and rem=1
now divide 8/3 gives r=2
now divide 85 successively by 2,3,5 and u will get rem 1,0,4
- 10 years agoHelpfull: Yes(5) No(0)
- First read the the question carefully.
follow these steps properly:
get started------
(According to the Divisibility rule DIVIDEND=DIVISOR*QUOTIENT+REMAINDER)
5(2(3k+2)+1)+0
5(6k+4)+1)+0
5(6k+5)+0
Now the number is 30k+25
according to the ques. we're dividing this no consecutively by 2,3,and 5 as below
(30k+25)/2= 15k+12 here when we divide 30k by 2 we'll get 0 as remainder but when we'll divide 25 by 2 we'll get 1 as remainder. OK, now next step we consider 15k+12 as the number instead of taking 30k+25.
now,
(15k+12)/3= 5k+4 as we discussed above the we'll get the remainder as 0.
Again, we're taking the no.
(5k+4)/5= k and 4 isn't divisible by 5 so it's considered as Remainder.
so finally we got the remainder is 1,0,4. - 9 years agoHelpfull: Yes(4) No(0)
- suppose quotient is K
we can say that from the given
5(2(3K+2)+1 -> 30K + 25 (dividend)
now divide successively 2,3 and 5 with the dividend.
we get 1, 0, 4 as remainder
- 10 years agoHelpfull: Yes(2) No(1)
- the number is 5 which gives remainder 0,1,2
nd the same number when divided by 2,3,5 remainder will come=1,2,0 ANS - 10 years agoHelpfull: Yes(2) No(2)
- answer is 1 2 0
Solution:- let us consider the number be 5. So according to calculation we get remainder as 1 2 0 when 5 is divided by 2,3,5. - 10 years agoHelpfull: Yes(1) No(0)
- 145 can be the no. as when divided by 5 r=0 ==> 29 by 2 r=1 ==>14 by 3 r=2
so when 145 is divided by 2 r=1 ==> 72 by 3 r=0 ==> 24 by 5 r=4
i.e
1,0,4 is the ans - 9 years agoHelpfull: Yes(1) No(0)
- This can be solvd through options easily...
- 9 years agoHelpfull: Yes(1) No(0)
- Answer is 1,2,0
if we consider the no is 5 then we get the remainder
5/5=R(0) , 5/2= R(1) , 5/3= R(2)
so
5/2= R(1), 5/3=R(2) , 5/5=R(0)
- 10 years agoHelpfull: Yes(0) No(0)
- that no can be.....35...65...95...and so on...because when divided by 5 it gives remainder 0.....so it is 5m type no....so remainder are....1...,,,2,,,and something for third no...
- 10 years agoHelpfull: Yes(0) No(0)
- let N be the number;
N=5*k;
k=2*t+1;
t=3*p+2;
p t k N
1 5 11 55
2 8 17 85
similarly we have more option.but
now try by taking N=55
55%5=0;
11%2=1
5%3=2;
so condition satisfied
now
55%2=1;
27%3=0;
9%5=4 - 9 years agoHelpfull: Yes(0) No(1)
- the remainder will be 1 in each case because that no. is obviously an odd no. then only you can get 0 remainder on dividing by 5 and 1 remainder on successively dividing by 2.
So a no. which is odd on dividing by 2 will give remainder 1 and after that if you will divide 1 by any natural no. (except 1 itself) you will get 1 as remainder. - 9 years agoHelpfull: Yes(0) No(0)
- remainder will be 1,0,5
- 9 years agoHelpfull: Yes(0) No(0)
- 1,2,0 is ans as the no is 5
- 9 years agoHelpfull: Yes(0) No(0)
- no is 65
and remainders are 1 2 0 - 9 years agoHelpfull: Yes(0) No(0)
- the no. is 30t+25 let t=(1,2,3........infinity)
- 8 years agoHelpfull: Yes(0) No(0)
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