If the 5-digit number 875ab is divisible by 3, 7 and 11, what is the sum of the reciprocals of a and b?
Answers
Step-by-step explanation:
Divisibility law of 3
⇒ A number is divisible by 3 if sum of its digit is divisible by 3.
Divisibility law of 11
⇒ If the difference of the alternating sum of digits of the number is a multiple of 11
(e.g. 2343 is divisible by 11 because 2 - 3 + 4 - 3 = 0, which is a multiple of 11).
(e.g. 2343 is divisible by 11 because (2 + 4) - (3 + 3) = 0, which is a multiple of 11).
Calculations:
If 5 digit number 535ab is divisible by 3, 7 and 11, then we can say number also divisible by 231 because = (3 × 7 × 11= 231).
From Divisibility law of 3 :
535ab is divisible by 3 if sum of its digit divisible by 3, so
5 + 3 + 5 + a + b
13 + a + b
As we know, (13 + a + b) is divisibly by 3 such that it can lie between minimum 15 and maximum 30.
⇒15 ≤ 13 + a + b ≤ 30
⇒ 2 ≤ a + b ≤ 17
⇒ 2 ≤ a + b ≤ 17 ----(i)
From Divisibility law of 11 :
5 digit number 535ab is divisible by 11 so definition from the divisibility law by 11
5 – 3 + 5 – a + b = 0
(5 + 5 + b) – (3 + a) = 0
⇒ (10 + b) – (3 + a) = 0
⇒ (b – a) = -7 ----(ii)
such that a > b
Possible cases:
a 9 8 7
b 2 1 0
Possible numbers are
1. 53592 ⇒ sum of digits is 24 it is divisible by 3
2. 53581 ⇒ sum of digits is 22 it is not divisible by 3
3. 53570 ⇒ sum of digits is 20 it is not divisible by 3
So, 53592 is the only possible number that is divisible by 3.
Also by checking divisibility of 53592 is also divisible by 7
So,53592 is the only possible number which is divisible by 3, 7 and 11
∴ a = 9 and b = 2.
⇒ (a2 - b2 + ab) = (92 - 22 + 9 × 2)
⇒ (a2 - b2 + ab) = 81 - 4 + 18
∴ (a2 - b2 + ab) = 95
If 5 digit number 535ab is divisible by 3, 7 and 11, then we can say number also divisible by 231 because = (3 × 7 × 11= 231).
let assume the largest number be 53599.
Now on dividing 53599 by 231
we get approx 232.92
So, the actual number = 231 × 232 = 53592
∴ a = 9 and b = 2.
⇒ (a2 - b2 + ab) = (92 - 22 + 9 × 2)
⇒ (a2 - b2 + ab) = 81 - 4 + 18
∴ (a2 - b2 + ab) = 95
Divisibility Test for 7
To Find out if a number is divisible by 7 or not, follow these steps:
1. Separate the last digit from the rest of the number. Let us call the rest of the number the truncated number. The truncated number has one less
digit than the original number or the previous truncated number.
2. Double the last digit and subtract it from the truncated number.
3. Check if this result is sufficiently small so that you can immediately say if this is divisible by 7. If it is divisible by 7, then so was the original
number. If it is not divisible by 7, then neither was the original number.
4. If the number is still too large to visually check if it is divisible, apply this rule over and over again as necessary.
E.g.Check 6132.
The last digit is 2 and the truncated number is 613.
Twice of 2 is 4. So subtract 4 from the truncated number 613 i.e. 613 – 4 = 609.
Again, the last digit is now 9, and the truncated number is 60.
Twice of 9 is 18. Subtract it from the truncated number 60, i.e. 60 – 18 = 42.
Now 42 is small enough to check visually.
We know that 42 is divisible by 7, so we can tell that 6132 is divisible by 7 also.
Solution :-
Case 1) :- 875ab is divisible by 3 .
- If sum of digits is divisible by 3 , the given number is also divisible by 3 .
So,
→ (8 + 7 + 5 + a + b) ÷ 3 = Remainder 0
→ (20 + a + b) ÷ 3 = Remainder 0
then, minimum value of a and b can be 0 and maximum value can be 9 .
taking minimum value ,
→ (20 + 0 + 0) = 20
taking maximum value,
→ (20 + 9 + 9) = 38
therefore, multiples of 3 between 20 and 38 are :-
- 21, 24, 27, 30, 33 and 36 .
- when sum = 21 => a + b = 21 - 20 = 1
- when sum = 24 => a + b = 24 - 20 = 4
- when sum = 27 => a + b = 27 - 20 = 7
- when sum = 30 => a + b = 30 - 20 = 10
- when sum = 33 => a + b = 33 - 20 = 13
- when sum = 36 => a + b = 36 - 20 = 16
Case 2) :- 875ab is divisible by 7 .
- Double the last digit number and subtract from rest part . If it is divisible by 7 , the given number also divisible by 7 .
So,
→ (875a - 2b) ÷ 7 = Remainder 0
Case 3) :- 875ab is divisible by 11 .
- When difference between sum of even places number and odd place number is 0 or divisible by 11 , the given number also divisible by 11 .
So,
→ (8 + 5 + b) - (7 + a) ÷ 11 = Remainder 0
→ (13 + b) - (7 + a) ÷ 11 = Remainder 0
→ (13 + b - 7 - a) ÷ 11 = 11 = Remainder 0
→ (6 + b - a) ÷ 11 = Remainder 0
→ (6 + b - a) = 0 or 11
→ (b - a) = (-6) or 5 .
Taking (b - a) = 5 :-
- if b = 5, a = 0 then a + b = 5
- if b = 6, a = 1 then a + b = 7
- if b = 7, a = 2 then a + b = 9
- if b = 8, a = 3 then a + b = 11
- if b = 9, a = 4 then a + b = 13
Therefore, from conclusion of divisibility by 3 and divisibility by 11 we can conclude that,
→ a + b = 7 where a = 1 and b = 6 .
Putting these in divisibility by 7 case we get,
→ (875a - 2b) ÷ 7 = Remainder 0
→ (8751 - 2 * 6) ÷ 7 = Remainder 0
→ (8751 - 12) ÷ 7 = Remainder 0
→ 8739 ÷ 7 = Remainder 0
again,
→ (873 - 9 * 2) ÷ 7 = Remainder 0
→ (873 - 18) ÷ 7 = Remainder 0
→ 855 ÷ 7 = Remainder 0
again,
→ (85 - 5 * 2) ÷ 7 = Remainder 0
→ (85 - 10) ÷ 7 = Remainder 0
→ 75 ÷ 7 = Remainder 0
Since 75 is not divisible by 7 . Therefore, value of a = 1 and b = 6 in not possible .
Now, taking (b - a) = 6 :-
since difference between (b - a) is negative, we can conclude that, a is greater than b .
Then, possible values of a and b are :-
- when b = 0 , a = 6 , then (a + b) = 6 + 0 = 6
- when b = 1 , a = 7 , then (a + b) = 1 + 7 = 8
- when b = 2 , a = 8 , then (a + b) = 2 + 8 = 10
- when b = 3 , a = 10 , which is not possible since a is a one digit number .
Therefore, from conclusion of divisibility by 3 and divisibility by 11 we can conclude that,
→ a + b = 10 where a = 2 and b = 8 .
Putting these in divisibility by 7 case we get,
→ (875a - 2b) ÷ 7 = Remainder 0
→ (8752 - 2 * 8) ÷ 7 = Remainder 0
→ (8752 - 16) ÷ 7 = Remainder 0
→ 8736 ÷ 7 = Remainder 0
again,
→ (873 - 6 * 2) ÷ 7 = Remainder 0
→ (873 - 12) ÷ 7 = Remainder 0
→ 861 ÷ 7 = Remainder 0
again,
→ (86 - 1 * 2) ÷ 7 = Remainder 0
→ (86 - 2) ÷ 7 = Remainder 0
→ 84 ÷ 7 = Remainder 0
we can check again,
→ (8 - 4 * 2) ÷ 7 = Remainder 0
→ (8 - 8) ÷ 7 = Remainder 0
→ 0 ÷ 7 = ÷ 7 = Remainder 0 (Satisfy .)
As we can see that, all three divisibility rule satisfy here . Therefore, a = 2 and b = 8 are correct values .
Hence,
→ 1/a + 1/b
→ 1/2 + 1/8
→ (4 + 1)/8
→ (5/8) (Ans.)
∴ The sum of the reciprocals of a and b is equal to (5/8) .
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