maths divisibility test
TRANSCRIPT
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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