Maths Divisibility Test

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    SUCCESS POINT SCIENCE ACADEMY,

    BHAKTI DHAM ROAD ,

    CHANDUR BAZAR

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    Prime Numbers

    Definition:If number other than 1 isdivisible by 1 and the number itself

    then the number is called Primenumber.

    If b and k are prime numbers then band k are prime factors of a.

    e.g. 210 = 2 x 3 x 5 x 7

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    2, 3, 5, 7 are the prime divisors ofgiven number.

    If x and y both are divisible by c thenx + y and x y are also divisible by c.

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    Tests of divisibility

    Divisibility test of 2, 4, 8: Consider numberabcd where a, b, c, d are digits.

    ThousandthHundredthTenth Unit

    a b c d

    1000a + 100b + 10c + d

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    Divisibility by 2, 4, 8

    If a number ends in a 0 or in an even numbersuch as 2,4,6 or 8, then it is divisible by 2.

    If a number ends in two zeros, or if thenumber formed by the tens and ones digit isdivisible by 4, then the number is divisible by4.

    If a number ends in three zeros, or if thenumber formed by the hundreds, tens andones digit is divisible by 8, then the number

    is divisible by 8.

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    Divisibility Test of 3 and 9

    Thousandth Hundredth Tenth Unit

    a b c d

    1000a + 100b + 10c + d

    (999+1)a + (99+1)b + (9+1)c + d

    999a + 99b + 9c + (a + b + c + d)

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    Divisibility test of 3 and 9

    If the sum of the digits of a number isdivisible by 3, the number is divisibleby 3.

    If the sum of the digits of a number isdivisible by 9 , the divisible by 9.

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    Divisibility test of 11

    a b c d e f g

    1000000a + 100000b+ 10000c+ 1000d +100e+ 10f + g

    (90909 x11+ 1)a + (909 x 11 + 11 -1)b + (909 x 11 +1)c + (90 x 11 + 11 1)d + (9 x 11 +1)e + (11 -1)f+g

    (90909 x 11) a +a + (909 x11 +11)b b + (909 x11)c+c + (90 x 11 +11)d d + (9 x 11)e + e + 11f f +g

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    = (90909)a + (909 x 11)b x (909 x 11)c + (90

    x11 +11)d + (9x 11)e + 11f + (a b + c d +e f + g)

    = (90909)a + (909 x 11)b x (909 x 11)c + (90x11 +11)d + (9x 11)e + 11f + [ a + c + e + gb d f ]

    Here a, c, e, g are digits at odd places andb, d, f are digits at even place.

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    Divisibility test of 11

    If the difference between the sum ofdigits in the odd number places(starting from the units place) and the

    sum of digits in the even-numberedplaces is either 0 or a multiple of 11,then number is divisible by 11.

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    Problems

    1. Test whether the given numbers aredivisible by number 3,6,9 and 11.

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    2. 7 * 6 * 8 is divisible by 11 then find

    numbers at *.

    3. How much minimum number should

    be subtracted from 8513 so thatremaining number will be divisible by 9?

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    Divisibility test of 7

    Consider three digit number abc.

    100a + 10b + c = (14 x 7 + 2)a + 10b + c

    = 14 x 7 x a + 2a + 10b +c

    In number abc, there are a times 100.

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    Test: Sum of twice of number of hundreds

    and number formed by remaining digits isdivisible by 7 then the number is divisibleby 7.

    e.g. consider numbers 819, 42875 and11482.

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    Divisibility test of 17

    Consider a three digit number abc.

    100a +10b + c

    = (17 x 6 -2)a +10b + c

    = 17x 6 x a 2 x a +10b + c

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    Test: If 2 times number of hundredssubtracted from number formed byremaining digits, if this difference isdivisible by 17 then the original number is

    divisible by 17.

    e.g. 31569

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    Divisibility test of 13

    Consider three digit number bcd.

    100b + 10c + d

    = (113 x 8 4)b + 10c +d

    = (113 x 8 x b) - 4b + 10c + d

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    Test: If difference obtained bysubtracting four times number ofhundreds from two digit number isdivisible by 13 then the number isdivisible by 13.

    e.g. 5369

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    Problem :

    Using 9 digits from 0 to 9, formsmallest 9 digit number divisible by

    11. Also find greatest 9 digit numberdivisible by 11.

    S l ti

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    Solution:1) 9 is greatest digit therefore using digit 0to 8 we can form smallest number as

    102345678.

    Apply divisibility test of 11

    8 + 6 + 4 + 2 + 1 = 210 + 3 +5 +7 = 15

    Thus decrease 21 by 3 and increase 15 by 3To satisfy test of 11 replace 5678 fromabove number by 7586.

    Answer 102347586

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    2. For greatest number neglect digit 0.

    987654321

    9 + 7 + 5 + 3 + 1 = 25 and 8 + 6 + 4 + 2 = 20

    To 25 - 20 = 5 to make this difference 0 ormultiple of 11 increase 25 by 3 anddecrease 20 by 3 therefore 28 17 =11 .

    To adjust this replace 4321 by 2413.

    Ans. 987652413

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    SUCCESS POINT SCIENCE ACADEMY,BHAKTI DHAM ROAD ,CHANDUR BAZAR

    DI- AMRAVATI