Find the least number being increased by unity will be exactly divisible by 22, 17, 33 and 34.
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Hi there !
Here's the answer :
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¶¶¶ TYPE OF PROBLEM :
To Find the least number being increased by k will be exactly divisible by x, y and z.
¶¶¶ APPROACH TO PROBLEM :
• Find LCM(x,y,z)
Since, The least Number divisible by x,y & x is the LCM(x,y,z)
• The Required least No. which when increased by k becomes exactly divisible by x, y & z = LCM(x,y,z) - k
¶¶¶ SOLUTION :
Given,
w, x, y & z = 22, 17, 33, 34
- Find LCM(22, 17, 33, 34):
2 | 22, 17, 33, 34
11 | 11 , 17, 33, 17
17| 1, 17, 3, 17
.• | 1, 1, 3, 1
•°• LCM(22, 17, 33, 34) = 1122
- Required Least No. = 1122 - 1 = 1121
Therefore,
The Number 1121 is the least Number which when increased by 1, gets exactly divisible by 22, 17, 33, 34.
•°•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°
Here's the answer :
•°•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°
¶¶¶ TYPE OF PROBLEM :
To Find the least number being increased by k will be exactly divisible by x, y and z.
¶¶¶ APPROACH TO PROBLEM :
• Find LCM(x,y,z)
Since, The least Number divisible by x,y & x is the LCM(x,y,z)
• The Required least No. which when increased by k becomes exactly divisible by x, y & z = LCM(x,y,z) - k
¶¶¶ SOLUTION :
Given,
w, x, y & z = 22, 17, 33, 34
- Find LCM(22, 17, 33, 34):
2 | 22, 17, 33, 34
11 | 11 , 17, 33, 17
17| 1, 17, 3, 17
.• | 1, 1, 3, 1
•°• LCM(22, 17, 33, 34) = 1122
- Required Least No. = 1122 - 1 = 1121
Therefore,
The Number 1121 is the least Number which when increased by 1, gets exactly divisible by 22, 17, 33, 34.
•°•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°
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Answer:
equal amounts of money are kept in three different boxes. the amount of the first box is distributed equally among 21, the amount of the second box equally among 24 women and that in the third box equally amount 32 . if the valence in every box be rs 9, what was the least amount in every box at first ?
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