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Ask QuestionPosted by Ayush Madan 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
The largest positive integer that will divide 398, 436 and 542 leaving reminders 7, 11 and 15 respectively is the HCF of the numbers (398 – 7), (436 – 11) and (542 – 15) i.e. 391, 425 and 527.
HCF of 391,425 and 527:
HCF of 425 and 391:
425 = 391 × 1 + 34
391 = 34 × 11 + 17
34 = 17 ×2 + 0
HCF of 425 and 391 = 17
527 = 17 × 31
Similarly, HCF of 17 and 527 = 17
So, HCF of (391,425 ,527) = 17
∴ Required number is 17.
Posted by Rishabh Joshi 6 years, 9 months ago
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Posted by Rishabh Joshi 6 years, 9 months ago
- 2 answers
Yogita Ingle 6 years, 9 months ago
x2 - 2x - 8
x2 - 4x + 2x - 8
= x (x - 4) + 2 ( x - 4)
= ( x + 2) (x - 4)
Ravi.....✌️ Ivar......? 6 years, 9 months ago
Posted by Disha Soni 6 years, 9 months ago
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Posted by Prashant Kumar Anand 6 years, 9 months ago
- 1 answers
Posted by Aanchal ????? 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0<r<3
i.e n = 3q -------- (1),
n = 3q +1 --------- (2),
n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.
Posted by Aanchal ????? 6 years, 9 months ago
- 1 answers
Posted by Aryan Raj 6 years, 9 months ago
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Posted by Sanjana Chimurkar 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
To find the largest number which exactly divides 280 and 1245 living remainders 4 and 3 respectively, we subtract 4 and 3 from 280 and 1245.
280 - 4 = 276
1245 - 3 = 1242
276 = 2 x 2 x 3 x 23
1242 = 2 x 3 x 3 x 3 x 23
HCF = 2 x 3 x 23 = 138
Therefore, the largest number which exactly devides 280 and 1245 living remainders 4 and 3 respectively is 138.
Posted by Aman Kumar 6 years, 9 months ago
- 3 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Aman Kumar 6 years, 9 months ago
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Posted by Suyash Kumar 6 years, 9 months ago
- 3 answers
Yogita Ingle 6 years, 9 months ago
2×3×15+7 = 6 ×15+7 = 90 + 7 = 97
97 is a prime number
So, 2×3×15+7 is prime number
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
अनुराग त्रिपाठी??? 6 years, 9 months ago
Posted by Kulwinder Singh 6 years, 9 months ago
- 1 answers
Posted by Abhi Singh 6 years, 9 months ago
- 6 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Pushpadevi Pushpadevi 6 years, 9 months ago
- 1 answers
Posted by Prasad Bhale 6 years, 9 months ago
- 1 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Hansin Sakalle 6 years, 9 months ago
- 0 answers
Posted by Aryan Raj 6 years, 9 months ago
- 2 answers
Yogita Ingle 6 years, 9 months ago
Last line is
Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 - 17 = 4663.
Yogita Ingle 6 years, 9 months ago
The given numbers are 520 and 468.
The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.
Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13
Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13
LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.
Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 × 17 = 4663.
Posted by Abhishek Shekhawat 6 years, 9 months ago
- 1 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by ?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
- 5 answers
Harsimran Singh™️ℹ️Ⓜ️ 6 years, 9 months ago
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by ?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
- 0 answers
Posted by Shagun Motla 6 years, 9 months ago
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Posted by Kaustubh Singh 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
According to the question, we have to show that every positive integer is either even or odd.
Let us assume that there exists a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n - 1 must be either odd or even.
Case 1: If n - 1 is even, n - 1 = 2k for some k.
But this implies n = 2k + 1
This implies n is odd.
Case 2: If n - 1 is odd, n - 1 = 2k + 1 for some k.
But this implies n = 2k + 2 = 2(k + 1)
This implies n is even.
Therefore,In both cases , we arrive at a contradiction.
Thus, every positive integer is either even or odd
Posted by Rupal Biranwar 6 years, 9 months ago
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Posted by Durgesh Kumar 6 years, 9 months ago
- 1 answers
Posted by Tejaswini Venuturumilli 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let a - d, a and a + d be the zeros of the polynomial F(x). Then,
Sum of the zeroes = {tex}- \frac { \text { Coefficient of } x ^ { 2 } } { \text { Coefficient of } x ^ { 3 } }{/tex}
{tex}\Rightarrow (a-d)+a+(a+d)=-\frac { ( - p ) } { 1 }{/tex}
{tex}\Rightarrow 3a=p{/tex}
{tex}\Rightarrow a=\frac { p } { 3 }{/tex}
Since {tex}a{/tex} is a zero of the polynomial {tex}f(x){/tex}. Therefore,
{tex}f(a)=0{/tex}
{tex}\Rightarrow a^3-pa^2+qa-r=0{/tex}
{tex}\Rightarrow \left( \frac { p } { 3 } \right) ^ { 3 } - p \left( \frac { p } { 3 } \right) ^ { 2 } + q \left( \frac { p } { 3 } \right) - r = 0{/tex}
{tex}\Rightarrow p^3-3p^3+9pq-27r=0{/tex}
{tex}\Rightarrow 2p^3-9pq+27r=0{/tex}
hence, {tex}2p^3-9pq+27r=0{/tex} is the required condition
Posted by Harshit Kumar 6 years, 9 months ago
- 2 answers
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Juhu Panda 6 years, 9 months ago
1Thank You