a. 0 b. 7896 c. 6993 d. 903
a. 7 b. 29 c. 41 d. 67
a. 1 b. 3 c. 7 d. 9
a. 2525 b. 2975 c. 3225 d. 3775
a. 2 b. 3 c. 6 d. 7
2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276
a. 5 b. 6 c. 7 d. None of these
a. 2 b. 5 c. 8 d. None of these.
264, 396, 462, 792, 968, 2178, 5184, 6336
a. 4 b. 5 c. 6 d. 7
a. divisible by 3 but not by 11 b. divisible by 11 but not by3 c. divisible by both 3 and 11 d. neither divisible by 3 nor by 11.
a. 6 b. 12 c. 24 d. 120
I. 2 II. 3 III. 5 IV. 6
a. Only I b. Only II c. Only I and II d. Only II and IV
a. 2 b. 3 c. 4 d. 5
a. 6 b. 12 c. 14 d. 18
a. 364724 b. 365387 c. 365737 d. 366757
a. 11 b. 13 c. 15 d. Data inadequate
a. 0 b. 1 c. 2 d. 3
a. 12 b. 24 c. 12-x d. multiple of 12
a. 214 b. 476 c. 954 d. 1908
a. 2 b. 3 c. 4 d. 5
a. 1220 b. 1250 c. 2 2030 d. 22030
Answer & Explanations
(Local value)-(Face value)= (7000-7)=6993.
x+(x+36)+y=100 => 2x+y=64
Therefore y must be even prime, which is 2.
Therefore 2x+2=64 => x=31
Third prime number= (x+36)= (31+36)= 67.
Unit digit in (3127)173 = Unit digit in 7173. Now, 74 gives unit digit 1.
Therefore, 7173= (74)43 * 71. Thus, 7173 gives unit digit 7.
(51+52+53+………+100) = (1+2+3+…….+100)- (1+2+3+4+……+50)
= (100*101)/2 - (50*51)/2
= (5050-1275)= 3775.
Let the number be 5b2. Clearly, it is divisible by 2.
Now, 5+b+2= (7+b) must be divisible by 3. So, b= 2.
Taking the sum of the digits, we have:
S1=9, S2=12, S3=18, S4= 9, S5=21, S6= 12, S7=18, S8=21, S9=15, S10=24.
Clearly S2, S5, S6, S8, S9, S10 are all divisible by 3 but not by 9. So, the number of required numbers= 6.
Since 4864 is divisible by 4, so 9P2 must be divisible by 3.
Therefore (11+P) must be divisible by 3.
Therefore least value of P is 1
A number is divisible by 132, if it is divisible by each one of 11, 3 and 4.
Clearly, 968 is not divisible by 3. None of 462 and 2178 is divisible by 4.
Also, 5184 is not divisible by 11.
Each one of remaining 4 is divisible by each one of 11, 3 and 4 and therefore, by 132.
Sum of digits= 35 and so it is not divisible by 3.
(Sum of digits at odd places)- (Sum of digits at even places)= 19-16=3, not divisible by 11.
So, the given number is neither divided by 3 nor by 11.
Required number= 1*2*3*4=24
Let the three consecutive odd numbers be (2x+1),(2x+3) and (2x+5). Their sum = (6x=9)= 3 (2x+3),which is always divisible by 3.
On dividing 6709 by 9, we get remainder= 4
Therefore, required number to be subtracted= 4
Required number is divisible by 72 as well as by 112, if it is divisible by their LCM, which is 1008.
Now, 1008 when divided by 72, gives quotient= 14.
Therefore, required number= 14.
Required number= 999*366+103= (1000-1)*366+103= 366000-366+103= 365737.
Number= (31 * Q)+ 29. Given data is inadequate.
Let x=6q+3. Then, x2= (6q+3)2= 36q2+36q+9= 6(6q2+6q+1)+3.
So, when 2n is divided by 4, remainder =3.
Putting x=2, we get 22(22-1)= 12. So, x2(x2-1) is always divisible by 12.
Z= 6*1 +4= 10.
Y= 5z+3= 53
X= 4y+2= 214.
Let n=3q+1 and let q= 2p+1. Then, n= 3(2p+1)+1= 6p+4
Therefore, the number when divided by 6, we get remainder= 4
Required number = (555+445)*2*110+30 = 220000+30= 220030.