Question

Q1. Show that one and only one out of n, n+4, n+8, n+12 and n+6 is divisible by 5 where n is any positive integer.

Q2. By using Euclid’s division algorithm , find the largest number which when divides 969 and 2059, gives the remainder 9& 11 respectively.

Q3. There are 156,208 and 260 students in groups A, B & C , respectively . Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired, if the same number of students should be accommodated in each bus.

Answer 1

Hey there !!

Any positive integer is of the form 5q , 5q + 1 , 5q + 2

here ,
b = 5
r = 0 , 1 , 2 , 3 , 4

when r = 0 , n = 5q
n  = 5q              ----> divisible by 5               ===> [1]
n + 4 = 5q + 4    [ not divisible by 5 ]
n + 8 = 5q + 8 [ not divisible by 5 ]
n + 6 = 5q + 6 [ not divisible by 5 ]
n + 12 = 5q + 12 [ not divisible by 5 ]

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when r = 1 , n = 5q + 1
n = 5q + 1  [ not divisible by 5 ]
n + 4 = 5q + 5 = 5 [q+ 1]    ----> divisible by 5 ===> [2]
n + 8 = 5q + 9 [ not divisible by 5 ]
n + 6 = 5q + 7 [ not divisible by 5 ]
n + 12 = 5q + 13 [ not divisible by 5 ]

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when r =  2 , n = 5q + 2
n = 5q + 2  [ not divisible by 5 ]
n + 4 = 5q +  6 [ not divisible by 5 ]
n + 8 = 5q +10 =  5 [q + 2 ]   --->  divisible by 5   ====> [3]
n + 6 = 5q +8 [ not divisible by 5 ]
n + 12 = 5q + 14 [ not divisible by 5 ]

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when r = 3 , n = 5q + 3
n  = 5q + 3   [ not divisible by 5 ]         
n + 4 = 5q + 7    [ not divisible by 5 ]
n + 8 = 5q + 11 [ not divisible by 5 ]
n + 6 = 5q + 9 [ not divisible by 5 ]
n + 12 = 5q + 15 = 5 [ q + 3 ]  --->  divisible by 5   ====> [4]
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when r = 4 , n = 5q + 4  
n  = 5q + 4   [ not divisible by 5 ]         
n + 4 = 5q + 8    [ not divisible by 5 ]
n + 8 = 5q + 12 [ not divisible by 5 ]
n + 6 = 5q + 10 = 5 [ q + 2 ]  --->  divisible by 5   ====> [5]
n + 12 = 5q + 16 [ not divisible by 5 ]

from 1 , 2 , 3 , 4  , 5 its clear that one and only one out of n, n+4, n+8, n+12 and n+6 is divisible by 5 
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q.2)
969 - 9 = 960
2059 - 11 = 2048

hcf of 960 and 2048 is the largest number that divides 969 and 2059 by leaving remainders .

960 < 2048
2048 = 960 × 2 + 128
960 = 128 × 7 + 64
128 = 64 × 2 + 0
hcf = 64

hence , 64 is the largest no:.

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to find the minimum no: of buses , with maximum no: of children in each bus , , lets first find  the  hcf of 165, 208 , 260

165 = 2 x 2 x 3 x 13
208 =  2 x 2 x  2 x 2  x 13
260 =  2 x 2  x 5 x 13

hcf = 2 x 2 x 13 = 52

maximum no: of students in 1 bus = 52

no: of students = 165 + 208  +  260 = 624

no: of buses = 624 / 52 = 12 

hence , 12 buses are required 

Answer 2

Q.1 change n+6 as n+16

step step explanation

by euclids division lemma

i hope it help you