Elitmus
Exam
Numerical Ability
Number System
A number p is 3 digit number like 1st digit is square of last digit. and no digit is same . so how many p is possible which is divisible by 3 .
Read Solution (Total 30)
-
- Three digit number = xyz
Rule1: 1st digit is square of last = x^2yx
Only numbers and their square are in single digit= 2,3 and their square are 4&9.
possible numbers:
4y2=402,412,432,452,462,472,482,492
9y3=903,913,923,943,953,963,973,983
Rule2: Number should be divisible by 3.
Now all possible numbers are:
4y2/3=402,432,462,492 = 4 count
9y3/3=903,963 = 2 count
hence total number count is 4+2=6. - 10 years agoHelpfull: Yes(85) No(3)
- total 6
903
963
402
432
462
492
- 10 years agoHelpfull: Yes(24) No(0)
- i dont think unit digit squares are only 2 & 3 even 1 have.(0,1,2,3,4,5,6,7,8,9)
1^2=111,141,171 which is 3 digit div by 3
2^2=402,432,462,492 which is div by 3
3^2=903,933,963,993
so total=11 - 10 years agoHelpfull: Yes(10) No(27)
- This is not for E-litmus level question.
- 10 years agoHelpfull: Yes(5) No(4)
- The three digit number is of the form a^2ba.
Now the numbers in the units digit can be (1,2,3) and in tens place can be (1,4,9) respectively.
Now when in units place 1 is there we get: 111, 141, 171 which are divisible by 3.
Now when in units place 2 is there we get: 402, 432, 462, 492 which are all divisible by 3.
Now when in units place 3 is there we get: 903, 933, 963, 993 which are again all divisible by 3.
Hence total of 11 numbers.. - 10 years agoHelpfull: Yes(4) No(5)
- ans. 6
913
923
943
953
973
983 - 10 years agoHelpfull: Yes(2) No(9)
- only 2 posible no. can take 1st position ie- 2 and 3with last digit being 4&9.
now to find the middle no combination, we know that for a no. divisible by 3.. sum of digit shud be divisible by 3.
So, 2+6 & 3+9 are already divisible by 3.
Now middle digit can be 0,3,6,9...
ie, no are 402,432,462,492,903,963.
so ans is 6 - 10 years agoHelpfull: Yes(2) No(0)
- total 8
903
933
993
963
402
432
462
492 - 10 years agoHelpfull: Yes(2) No(18)
- there are 3 possibilities for last digit.....
which is 1,2,3..
for digit 1.. 111,141,171...total 3 numbers..
for digit 2....402,432,462,492...total 4 numbers..
for digit 3..903,933,963,993...total 4 numbers..
answer is =3+4+4=11
- 9 years agoHelpfull: Yes(2) No(2)
- ans . 6
903
923
432
462
492
936
943 - 10 years agoHelpfull: Yes(1) No(6)
- only 6 no is possible
4m2,9n3
this is general no format sum of these no 4+m+2 will be divisible by 3
similarly others no
so only 402,432,462,492
903,963 - 10 years agoHelpfull: Yes(1) No(0)
- answer is 6
- 10 years agoHelpfull: Yes(1) No(0)
- The answer will be 11.
condition 1: 1st digit is square of last digit.
The possible 1st and last digit combination will be : 1st-1, last-1
1st-4, last-2
1st-9, last-3
condition 2: The number should divisible by 3.
Therefore, for 1st=1 and last=1, the possible number will be- 111, 141, 171.
for 1st=4 and last=2, the possible number will be- 402, 432, 462, 492.
for 1st=4 and last=2, the possible number will be- 903, 933, 963, 993.
Hence total number of combination will be: 3+4+4=9.
- 10 years agoHelpfull: Yes(1) No(5)
- Ans 5
432
462
492
903
963 - 10 years agoHelpfull: Yes(0) No(5)
- 804,834,864,894 can also satisfy
- 10 years agoHelpfull: Yes(0) No(11)
- ans is 6
4_2
9_3
are the possibilities.....( 402,432,462,492
903,963 ) - 10 years agoHelpfull: Yes(0) No(0)
- ans is 6
903, 963,402,432,462,492 - 10 years agoHelpfull: Yes(0) No(0)
- The correct answer is 6
let the number P=xyz
1st it is said that=z^2yz
Only numbers and their square are in single digit =1,2,3 and their square are 1,4,9
we will not take 1 as in question it is said that no digit is same.
Possible value:
2^2y2=402,,432,462,492 =4
3^2y2 =903,963=2
so total numbers are 6.
- 10 years agoHelpfull: Yes(0) No(0)
- Total is 8....
402,432,462,492
903,933,963,993 - 10 years agoHelpfull: Yes(0) No(1)
- 11
111,141,171
402,462,432,492
90,933,963,993 - 10 years agoHelpfull: Yes(0) No(0)
- 3 ways for no 4 _ 2 and 4 ways for 9 _ 3 which is divisible by 3. so,
total of 3+4 = 7 ways and 7 possible no is there with non repeating three digit no. - 9 years agoHelpfull: Yes(0) No(0)
- 6
402
432
462
492
903
963 - 9 years agoHelpfull: Yes(0) No(0)
- possible numbers are:- 402,432,462,492,903,963,
so total no can be possible only :- 6. - 9 years agoHelpfull: Yes(0) No(0)
- Rashmi or minupala boki h...
- 9 years agoHelpfull: Yes(0) No(0)
- 402
432
462
492
903
963 - 9 years agoHelpfull: Yes(0) No(0)
- 903
963
402
432
462
492
Total no of p=6 - 9 years agoHelpfull: Yes(0) No(0)
- No digit should be same..
4_2 :- 402,432,462,492..total 4
9_3 :- 903,963.. total 2
total number is 6 - 9 years agoHelpfull: Yes(0) No(0)
- right answer 8 number
402
432
462
493
903
933
963
993
and 111 not including because in question say no digit are same - 9 years agoHelpfull: Yes(0) No(0)
- 1^1=1(rejected.since the numbers in the digits cannot repeat itself according to the question)
2^2=4
3^2=9
so the 3 digit numbers will be- 4y2 & 9y3
check y=0,1,2....9 and check the numbers divisibility by 3 such that every digit in the 3 digit number is unique
the numbers divisible by 3 are: 402,432,462,492,903,963 are the required numbers.
- 9 years agoHelpfull: Yes(0) No(0)
- 402,432,462,492,903,963=6 numbers
- 9 years agoHelpfull: Yes(0) No(0)
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