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Posted by Vasu Singh 5 years, 7 months ago
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Posted by Om Jagtap 5 years, 7 months ago
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Akhil? . 5 years, 7 months ago
An = a + (n-1) d
=> 994 = 105 + (n-1) 7
=> 889 = 7n - 7
=> 896 /7 = n
=> n = 128
So, 128 three digits number are divisible by 7..
Posted by Meeinstitute Mdsmee Dance Studio 5 years, 7 months ago
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B.Aishwarya 203 Kalyannagar 5 years, 7 months ago
Akhil? . 5 years, 7 months ago
Lucifer?? Morningstar?? 5 years, 7 months ago
Posted by Ashok Yadav 5 years, 7 months ago
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Meghna Thapar 5 years, 7 months ago
Gachibowli is a suburb of Hyderabad, Telangana, India, located in the Serilingampally mandal of the Rangareddy district. It is situated about 5 km away from HITEC City, another IT hub. It has a vast area and is dotted with rocky surface and hillocks all around.
Posted by Amrita Kushwaha 5 years, 7 months ago
- 2 answers
Yogita Ingle 5 years, 7 months ago
Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are intezers.
so, √5 = p/q
p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesnt occurs with √5 since its not an intezer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.
? Royal Thakur ? 5 years, 7 months ago
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Posted by Vandana Raj 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
The prime factorization of 11008 is
2|11008
2|5504
2|2752
2|1376
2|688
2|344
2|172
2|86
43|43
1
11008=2×2×2×2×2×2×2×2×43
11008=2^8×43
The prime factorization of 7344 is
2|7344
2|3672
2|1836
2|918
3|459
3|153
3|51
17|17
1
7344=2×2×2×2×3×3×3×17
7344=2^4×3^3×17
Now, HCF=Product of the smallest power of each common
prime factor in the numbers.
HCF=2^4=16
LCM = Product of the greatest power of each prime factor,
involved in the numbers.
LCM=2^8×43×3^3×17=256×43×27×17=5052672
Posted by Bhavisha Singla 5 years, 7 months ago
- 1 answers
Yogita Ingle 5 years, 7 months ago
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 ]
-------------------------------------------
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 ]
----------------------------------------------
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 ]
----------------------------------------
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]
----------------------------------------------------
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
Posted by Ashok Yadav 5 years, 7 months ago
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Posted by Kailesh Aadhithyaa 5 years, 7 months ago
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Posted by Aditya Dinnimani 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
x2 - 21x+ 108
= x2 - 12x - 9x + 108
= x(x - 12) - 9(x - 12)
= (x - 12)( x -9)
Posted by Muskan Bano 5 years, 7 months ago
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Posted by Aryan Kumar 5 years, 7 months ago
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Dr. Arizoo 5 years, 7 months ago
Ayush Rai 5 years, 7 months ago
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