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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , 540 dks 2 lgvHkkt; la[;kvksa ds xq.ku ds :Ik esa, fdrus rjhds ls O;Dr dj ldrs gSA, , 8000 ds fdrus QSDVj iw.kZ ?ku la[;k gSA, (a) 12, , (b) 6, , (a) 7, , (b) 10, , (c) 8, , (d) 4, , (c) 4, , (d) 6, 25., , 20., , Find the number of such numbers that are less, than 250 and co-prime to 250., , 1600 ds fdrus vyx&vyx QSDVj iw.kZ oxZ gSA, , ,slh la[;vks dh la[;k tks 250 ls de gS o 250, dh lgvHkkT; gSA, , 21., , (a) 100, , (b) 200, , (c) 105, , (d) 205, 1, 6, 11, 16, 21, , How many divisors of 105 will have no zeros at, its end?, , 105 ds fdrus Hkktd viuk vfUre vad 'kqU; ugha, j[krs gSA, (a) 5, , (b) 11, , (c) 8, , (d) 9, , If n = p2×q3×r4, where, p, q and r are primes., Find the number of factors of N which are, perfect squares., , ;fn n = p2×q3×r4 tgkW p, q, r vHkkT; gSA N ds iw.kZ, oxZ QSDVjks dh la[;k Kkr djks\, , 23., , (a) 12, , (b) 14, , (c) 16, , (d) 8, , (c) 8, , (d) 3, , (c) 9, , (d) 8, , a, c, c, d, b, , 2, 7, 12, 17, 22, , a, b, a, d, a, , Answer Key, 3, a, 8, d, 13 a, 18 d, 23 c, , 4, 9, 14, 19, 24, , 5, 10, 15, 20, 25, , a, d, a, a, d, , (a), n=6, , n2 = 36, , 21×31, , 22×32, , Divide it by 2 to find the pairs of these factors =, 10/2 = 5, Now find the total number of factors of N =, (1+1)(1+1) = 4, So to find the factors of N2 which are less than, 6 and do not divide N completely, No. of pairs of factors of N2 – No. of total factor, of N ⇒ 5-4 = 1, OR, Factors of 36 ⇒ 22×32, , 24., , b, c, a, c, b, , Add 1 to the total factors to make is even, number = 9+1 = 10, , 630 ds le Hkktdks dh la[;k o vHkkT; Hkktdks dh, la[;k ds chp vUrj D;k gksxkA, (b) 24, , (b) 10, , So total No. of factors of N2 = (2+1)(2+1)= 9, , What is the difference between the number of, even divisors and the number of prime, divisors of 630?, , (a) 12, , (a) 12, , Solutions, 1., , 22., , How many distinct factors of 1600 are perfect, squares?, , How many distinct factors of 8000 are perfect, cubes?, , Factors are 1,2,3,4,6,9,12,18,36, , 3

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , So factors which are less than N i.e. 6 are, 1,2,3,4,6 and the factor which do not, completely divides the n (i.e.6) is only one, factor which is 4., , 4., , (b), Prime factors of 35×72×152×43 ⇒ 26×37×52×72, Prime factors of 1260, , = 22×32×5×7, , So let k be the multiple of 1260, 2., , (a), , So we can = k(22×32×5×7) = 26×37×52×72, , N2 = 330×739, , So K = 24×35×51×71 are the multiple of, 24×35×5×7 out of 26×37×52×72, , N2 = 318×778, , ⇒ i.e. factors of 24×35×51×71, , Total factors of N2 = 19×79 = 1501, , ⇒ 5×6×2×2 = 120, , Add +1 to make it even = 1501+1 = 1502, No. of pairs = 1502/2 = 751, 5., , No. of factors in N = 10×40 = 400, , (a), Prime factorization of 360, , Desired no of factors = 751-400 = 351, , This is should be in the form of 2a×3b×5c, 3., , So product of the factors of 360 is, , (a), , 432, ,  N12, , Prime factors of 60 = 22×31×51, , N 2, , Here all the factors that can be written as, multiple of 60 will be of the form., , i.e (360)12, , 22×31×51×K, , 6., , So 210×36×511 = K(22×31×51), , (c), Prime factorization of 60 = 22×3×5, , K = 28×35×510, , So the No. of should be of form = 2a×3b×5c, , Now No. of factors of 28×35×510 is, (8+1)(5+1)(10+1), , Total No. of factors is = (a+1)(b+1)(c+1), , = 9×6×11 = 594, , = 3×2×2 = 12, , OR, , 12, , Here we are looking for factors of, that are multiple of 60. These will also to be of, form 2p×3q×5r. But as 60 is 22×31×51, the set of, value p, q, r can take are limited, , Now the product of factors, , 210×36×511, , N 2, ,  , ,  N6  606, , 7., , (b), , values can be taken by p = 10-2 = 8, , Prime factorization of 180 = 22×32×5, , q = 6-1 = 5, , This should be of form, , = 2a×3b×5c, , r = 11-1 = 10, , No. of factors of 90, 3×3×2 = 18, , = (a+1)(b+1)(c+1) =, , So the number of factors that are multiples of, 60 will be 9×6×11 = 594, , Now, product of the factors of 180 is

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , N = 24×32×118 → This should be of form =, 2a×3b×11c, , 18, , 180 2, , =, , 1809, , Total factors of N= (a+1)(b+1)(c+1) ⇒, 8., , (d), , (a+1)(b+1)(c+1)  3×2×5 = 30, , Prime factorization of 900 = 22×32×52, , Total factors of N2 = 5×3×9 = 135, , This should be of form = 29×32×52, , Add 1 to make it even number = 135+1 = 136, , a can take values = 0, 2, , No. of pairs , , b can take values = 0, 2, , 136,  68, 2, , Desired number of factors = 68-30 = 38, , c can take values = 0, 2, So the possibilities ⇒, 11., , a b c, 0, 0, 0, 0, , 0, 0, 2, 2, , 2, 2, 2, 2, , 0, 2, 0, 2, , 0, 2, 0, Sothere are total, 2, 8 factors that areperfect squares., 0, 0, 2, 2, , (c), Three digit numbers divisible by 5 or 11 Three, digit numbers divisible by 5+ three digit, numbers divisible by 11- three digit numbers, Prime, factorization, of 300 = 22×3×52, divisible, by 5 and 11., Three digit numbers divisible by 5  100,, 105…..995, There are in AP with d = 5, 995 = 100+(n-1)×5, , 9., , (c), This should be of form = 2a×3b×5c, , Three digit numbers divisible by 11 = 110, 121,, ……990, , a can take values = 0, 2, , There are in AP with d = 11, , b can take values = 0, , 990, , c can take values = 0, 2, , Three digit number divisible by 55 =, , So, possibilities are, , 990, , a b c, 0 0 0, 0 0 2 total4possibilities, , So three digit number divisible by 5 or 11, , = 110+(n-1)11  n = 81, , = 110+(n-1)55  n = 17, , = 180 + 81-17 = 244, , 2 0 0, 2 0 2, , 10., ,  n = 180, , 12., , (a), Two digit numbers divisible by 2 or 3 = Two, digit numbers divisible by 2+ two digit, numbers divisible by 3-two digit numbers, divisible by 2 and 3, , (d), N = 22×31×114, ,  Two digit numbers divisible by 2., , 5

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , SO we have one way A choosing 2, 1 way of, choosing 3 and 3 ways of choosing 5 Thus, required number of factors are 1×1×3 = 3, , 98 = 10 + (n-1) 2  n = 45,  Two digit numbers divisible by 3., 99 = 12 + (n-1) 3  n = 30,  Two digit numbers divisible by 6., , 15., , a = 12, an = 96, d = 6, , We have seen that the number of co primes to, 60 upto 60 = 16, , 96 = 12 +(n-1)×6  n = 15, , Hence, sum of co primes to 60 upto 60 =, , So the number divisible by 2 or 3 are = 45 + 30, -15 = 60, , 13., , (a), , 16,  60  480, 2, , (a), , 16., , If 814A51B is divisible by 88 then it must, divisible by 8 and 11 also., , (d), 100 = 22×52, Number of factors = (2+1)(2+1)= 9., , So by divisibility of 11, Number of ways =, , 17+B = 2+A  A-B = 15 or A = 15+B, , 9 1, 4, 2, , …..(1), , Now, by divisibility of 8, 17., , 51 B should by divisible by 8 so B can take only, one value i.e.[2]-, , (d), 32100 = 22×3×52×107, Number of prime pairs = 24-1, , So from 1 = A = 15+2 = 17 or = 17-A this only, possible when A is [6] i.e., , = 23 =8, , 17-A is zero or multiple of 11, 18., , So the number is, , 840 = 23×3×5×7, , 8146512, So A+B, And A×B, , 14., , (d), , Possible number co prime to 840 upto 840 =, , = 6+2 = 8, , 1 , 1 , 1 , 1, ,  840   1   1   1   1  , 2 , 3 , 5 , 7, , 1 2 4 6,  840    , 2 3 5 7,  192, , = 6×2 = 12, , (a), Prime factors of 7200  25×32×52, , Sum of all such pairs =, , Since we are taking even factors so there must, be at least one 2 in required factors. And the, number is divisible by 9 so we must take both, threes., , 192,  840, 2, , = 96×840=80640, , 19., , We can’t take more than 1 two as it will make, the number divisible by 36., , (c), 540 = 22×22×5, Multiplication of 2 co-prime number factor, , 6

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Edited with the trial version of, Foxit Advanced PDF Editor, To remove this notice, visit:, www.foxitsoftware.com/shopping, , (1, 540) (4, 135) (7, 20) (5, 180), , 2 can have value = 20, 23, 26, , 4 pair is possible, , 5 can have value = 50, 53, So total perfect cube factor, , 20., , (a), , = 3×2 = 6, , 250 = 2×53, , 21., , 25., , Co-prime of 250 less than 250, , 1600= 26×52, , 1 , 1, ,  250  1   1  , 2 , 5, , 1 4,  250  , 2 5,  100, , Here, , So total perfect square distinct factor = 4×2=8, , (b), , Number of factor that do not have 5 are = 5, Factor that don't have zero they also not have, 2 as its factor so number of factor without 2, are, = (5+1) = 6, So total number of factor without unit digit, zero = 6 + 5 = 11, , (a), perfect square factor, = (p0+p2)(q0+20) (r0+r2+r4), Total number of perfect square factor, = 2×2×3, = 12, , 23., , (c), 630 = 2×32×5×7, Even number of factor = 1×3×2×2 = 12, Number of prime factor = 4, Difference = 18 – 4 = 8, , 24., , 2 can be 20, 22, 24, 26 = 4 value, 5 can be 50, 52 = 2 value, , 105 = 25×55, , 22., , (d), , (b), 800 =26×53, , 7