18/08/2018 7:36 am
Number system
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Sir Help!
Find the digit A if the number 888….888A999….999 is divisible by 7, where both the digits 8 and 9 are 50 in number.
This post was modified 6 years ago by Richa Malhotra
TG Team 18/08/2018 11:06 am
Hello Richa ,
Any digit repeated (p-1) times is always divisible by p if p is a prime number greater than 5.
So 888888 written 6 times is divisible by 7
[ or 888888 = 888 × ( 1001) = 888 × 7 × 11 × 13 so any digit repeated 6 times will be divisible by 7 , 11 and 13. ]
Hence , 88888.... written 48 times ( a mutiple of 6) and 9999.... written 48 times will be divisible by 7
So we need to find the value of A for which 88A99 is divisible by 7 .
88 × 1000 + A99 should be divisible by 7
-4 + A × 100 + 99 should be divisible by 7
-4 + 2A + 1 should be divisble by 7
A = 5
0
04/09/2018 10:41 pm
TG Team 04/09/2018 11:30 pm
Hi Tarishi
We Know,
an + bn is divisible by a + b when n is ODD.
an – bn is divisible by a + b when n is EVEN.
an – bn is ALWAYS divisible by a – b.
So ,
182000 + 122000 – 52000 -12000
182000 – 52000 is divisible by (18 -5) i.e 13
122000 - 12000 is divisible by (12 + 1) i.e 13
so the whole expression is divisible by 13.
Again ,
182000 –12000 is divisible by ( 18 – 1) i.e 17
122000 – 52000 is divisible by ( 12 + 5) i.e 17
So the whole expression is divisible by 17.
Hence 182000 + 122000 – 52000 -12000 is divisible by ( 17 x 13) i.e 221.
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05/09/2018 12:05 am
let b be a positive integer and a =b^2-b . If b >= 4 , then a^2-2a is divisible by :
a) 15
b)20
c) 24
d) All of the above
This post was modified 5 months ago by TathaGat
TG Team 05/09/2018 3:25 pm
a = b2 – b
a2 – 2a = a (a – 2)
= (b2 – b)( b2 – b – 2)
= (b -1)(b-2) b (b+1)
Product of four consecutive integers is always divisible by 4! i.e = 24 .
hence Option C .
0
08/09/2018 10:26 pm
What is the remainder when 7777777.....201 digits is divisible by 41? Can we solve it using Euler's theorem?
TG.Raman 09/09/2018 12:35 am
Hi Aarushi ,
Any digit repeated ( p -1) times is always divisible by p if p is a prime number greater than 5. ( an application of Euler's theorem )
So 77777..... written 40 times is divisible by 41.
Hence 77777.... Written 200 times is a multiple of 41.
Net remainder : 7
0
02/10/2018 12:51 pm
TG Team 02/10/2018 1:59 pm
Common factors means Factors of HCF ( P,Q)
HCF (P,Q) = 27 x 35 x 73
For perfect squares , we can select :
20, 22 , 24 , 26 in 4 ways
30 , 32 , 34 in 3 ways
70 , 72 in 2 ways
We need even perfect square so at least one 2 should be there
Hence , 3 x 3 x 2 = 18 factors ..
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02/10/2018 12:53 pm
TG Team 02/10/2018 1:31 pm
Let the three consecutive odd integers be a-2 , a and a + 2
(a -2)2 + a2 + (a +2)2 = 3a2 + 8
Only option C gives remainder 2 when divided by 3 Hence , 5555.
0
10/10/2018 11:04 am
How many 18-digit positive integers are there which ends in 18 as last two digits and are divisible by 18?
TG Team 10/10/2018 1:44 pm
18 = 2 x 9 so the number has to be divisible by 2 and 9 both.
From the question it is clear that the number ends with 18 , hence it is divisible by 2 .
Sum of remaining digits must be a multiple of 9 .
Hence all 16 digit numbers divisible by 9 should be the answer.
From 1015 to 1016 -1 there are [1016 – 1015]/9 = 1015 multiples of 9.
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12/10/2018 4:55 pm
| A and B when divided by 56 leave remainders of 48 and 32 respectively. When divided by 44, both leave a remainder of 24. If the sum of A and B is divisible by the sum of the divisors, then find the minimum value of (A + B). |
aniket prajapati 12/10/2018 7:23 pm
This post was modified 6 years ago by aniket prajapati
A=56a+ 48
B=56b+ 32
A+B= {56(a+b) +80}
A and B leave reminder 24 when divided by 44
A+B will leave reminder: (24+24)/44 = 4
A+B={ 56(a+b)}/44= 4(reminder)
Check value of a+b by putting 1,2,3.....
a+b =1 satisfies here and it is minimum hence
A+B= 56*(1)+80
=136 (answer)
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14/10/2018 12:05 am
Nine distinct digits appear in the product of 2, 7,181, 241 and 607. Which digit is missing?
(a)1 (b)2 (c)4 (d)6
TG Team 14/10/2018 1:17 pm
The digital sum of a number N is same as the remainder obtained by dividing the number N by 9.
Digital sum of the product 2 x 7 x 181 x 241 x 607 is 2 x 7 x 1 x 7 x 4 = 28 x 14 => 5 x 1 ;i.e 5.
Since the sum of all digits from 0 to 9 (1234567890) is 45, which is divisible by 9.
The only digit missing that could produce a remainder of 5 by 9 when divided by 9 is 4.
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15/10/2018 5:50 pm
How many natural numbers from (111)6 to (1111)6 are divisible by 9?
TG Team 15/10/2018 5:57 pm
Decimal representation of 1116 = 62 x 1 + 6 x 1 + 1 = 43
Decimal representation of 11116 = 63x1 + 62x 1 + 6 x 1+ 1= 259 .
Now , count multiples of 9 between these two numbers : [ 252 - 45]/2 + 1 = 24 Numbers .
0
17/10/2018 12:49 am
Let the product of first thousand even positive integers is A and product of first thousand odd positive integers is B. Find the remainder when A – B is divided by 2001?
TG Team 17/10/2018 1:28 am
Product of first 1000 even numbers : 2 x 4 x 6 x 8 x........x 2000
= 2^1000 x ( 1 x 2 x 3 x ..... x 665 x 666 x 667 x ........ x 999 x 1000) = A
Product of first 1000 odd numbers : 1 x 3 x 5 x 7x .... 665 x 666 x 667 ... x 1999 = B
2001 = 3 x 667
A and B both the numbers are divisible by 3 and 667 .
hence net remainder 0.
0
17/10/2018 12:49 am
How many of the following numbers are prime?
I. 172012 – 19
II. 152013 – 13
III. 142031 – 27
TG Team 17/10/2018 1:15 am
I. 172012 – 19 => difference of two odd number is always even so composite
II. 152013 – 13 => even , divisible by 2 Hence , composite
III. 142031 – 27 => (14677 )3 – 33
( A3 - B3) is always divisible by (A - B), so 142031 – 27 is a composite number.
Hence , None
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18/10/2018 4:56 pm
if n=539*2^18 and m=9*2^13 , then the remainder when n is divided by m is?
0
21/10/2018 12:57 am
Hi sir, please guide
0
21/10/2018 1:05 am
TG Team 21/10/2018 9:48 am
77777...... 1001 times
Any digit repeated 6 times is always divisible by 1001 so
77777.... Written 996 times is a multiple of 1001
Hence , 77777 /1001=>
777 -77 = 700