From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.
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In a series of numbers , the next number is formed by adding 1 to the sum of the previous numbers, and the 10th number is 1280. Then what is the first number in the series? (series will be like this x , x+1, (x+(x+1))+1,....... )
The given series is x, x + 1, 2x + 2, 4x + 4 ......
If you observe the pattern here, the coefficient of x + 1 is in the powers of 2. So 4th term has a power of 2, 5th term has a power of 3... 10th term has a power of 8. So tenth term would be 2^8(x + 1)
= 256(x+1).
Given 256(x+1) = 1280
x = 4.
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The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
We can write 10 = 1 + 2 + 3 + 4. So we have to find how many multiples of 10 can be written in this manner.
Let the first of the four numbers be n. So
n + (n+1) + (n+2) + (n+3) = 10k
4n + 6 = 10k
2n + 3 = 5k
n = (5k-3)/2
=2k-1+(k-1)/2
So n is integer for k = an odd number. So for k = 1, 3, 5, .... 99 we can write a number as a sum of four consecutive intezers.
So there are 50 numbers.
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Mr. Bean chooses a number and he keeps on doubling the number followed by subtracting one from it, if he chooses 3 as initial number and he repeats the operation for 30 times then what is the final result?
Let the number be N.
So N = 406x + 115.
Now divide this number by 29. As 406 is exactly divisible by 29, we have to divide 115 by 29 and find the remainder. So remainder = 28
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Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?
Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
=> 15x + 1 = 7c + 6 => c = (15x?5)/7 => c = 2x+(x?5)/7
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N - M = 56
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The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573. What is the largest of the numbers A, B, C, D?
a+b+c=4024
b+c+d= 4087
a+c+d=4524
a+b+d=4573
Combining all we get 3(a+b+c+d) = 17208
=> a + b + c +d = 3736
Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.
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In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?
Lets assume number of engg = e
number of Mba = m
and Number of CA = c
4e+3m+5c=3650
3c=2m m=3c/2
3e=2c e=2c/3 ------------- (1)
put the value in the above eq. (1)
4*2c/3+3*3c/2+5c=3650
c=300
m=3c/2
m=450
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Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460?
5678 - 35 + (one of the answer option) should be divisible by 460. Only option C satisfies.
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A number when successively divided by 5.3.2 gives remainders of 0, 2 and 1 respectively in that order. What will be the remainder when the same number is divided successively by 2, 3 and 5 in that order?
Let the number = N
Now N = 5K
K = 3L + 2
L = 2M + 1
K = 3(2M + 1) + 2 = 6M + 5
N = 5(6M + 5) = 30 M + 25
For M = 0 we get the least number as 25.
Now when 25 is divided by 2, we get 12 as quotient and 1 as remainder.
When 12 is divided by 3 we get 4 as quotient, and 0 as remainder. When 4 is divided by 5 we gt 4 as remainder.
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