M16. The product of a two-digit number by a number consisting of the same digits written in the reverse order is equal to 2430. Find the lower number? 45 M17. For each of the numbers: 41, 83, 32, the first digit is greater in value than the second digit.How many 2-digit numbers have this property? M18. Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor? M19.How many 3-digit numbers exist for which the sum of the digits is six? M20. Find the smallest number, greater than 1, which has a remainder of 1 when divided by any of 2, 3, 4, 5, 6, or 7 M21. 165 + 215 is divisible by 33 , 13 , 27 or 31? M22. How many pairs of natural numbers are there the difference of whose squares is 45? M23. Given that a, b, and c are natural numbers, solve the following equation. a!b! = a! + b! + c2 M24. A five digit number 3A25B is divisible by 19 and 7. Find A and B? M25. How many keystrokes are needed to type numbers from 1 to 100? M26. How many 3-digit numbers have two digits the same? M27. 14n + 11 can never be a M28. The remainder obtained when 43101 + 23101 is divided by 66? M29. What is the least number that should be multiplied to 100! to make it perfectly divisible by 350 M30. How many times will the digit '0' appear between 1 and 10,000? END |
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M16 - 45
M17 - 38 (countr the numbers between 10 and 99 which satisfy the mentioned ppty)
Abi
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Not sure.. but the answer shud be 599
Solution:
Ppty: X mod Y + Z mod Y = (X + Z) mod Y = (X mod Y + Z mod Y) mod Y
(431 + 379 ) mod y = 810 mod y = 211
Hence divisor = 810-211 = 599
Abi
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Ans : 21 numbers
Abi
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Solution:
LCM(2,3,4,5,6,7) + 1 = 840 + 1 = 841
Abi
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Slight change to the ppty but the answe still holds good
Ppty: X mod Y + Z mod Y = (X mod Y + Z mod Y) mod Y
Abi
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Dont know the shortcut method..
But, the answer is 13
Abi
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Dont know any proper method, but when u have choices this questionc an easily be done by substitution method.
Ans is a = 2 or 3, b = 2 or 3, c = 4
Abi
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102 key strokes
Abi
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C ... it cannot be even
Abi
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Excellent question ......... The number to be multiplied is 9...
Solution:
No of 3's in 100 = 33
No of 3^2 's in 100 = 11
No of 3^3 's in 100 = 3
No of 3^4 in 100 = 1
Now... 3^5 is greater than 100 and hence we can stop with 3^ 4.
This means that 100! already has 48 3's in it.
When 100! / 3^ 50, the 48 3's in 100! gets cancelled out with 48 3's in 3 ^ 50. Hence 9 is left over in the denominator.
Hence for 100! to be exactly divisible by 9, it as to be multiplied to 100!.
Abi
n/a
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