Find the least number whin when divided by 20,30&40 leave a remainder of 7? Please show working steps
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Solution:
To find: the least number when divided by 20,30&40 leave remainder 7,.
we can find the number by finding LCM of these numbers(as when divided by them,will give the least number),and by adding 7 to the resulting number,.
LCM{20,30,40}=120,.
120+7=127
∴The least number when divided by 20,30& 40 which leave remainder 7 is 127. Hope it Helps.
To find: the least number when divided by 20,30&40 leave remainder 7,.
we can find the number by finding LCM of these numbers(as when divided by them,will give the least number),and by adding 7 to the resulting number,.
LCM{20,30,40}=120,.
120+7=127
∴The least number when divided by 20,30& 40 which leave remainder 7 is 127. Hope it Helps.
sivaprasath:
Mark as brainliest,.
127/20 leaves 7, 127/30 leaves 7,127/40 leaves remainder 7
thank you
bro,mark as brainliest(then,It would be a thanks that is given to me.If it really helped)
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Sol)Let us see the relation between the pairs of numbers if there is any:
20-7=13
30-7=23
40-7=33
Let us add all these numbers 13+23+33=69
Next find the LCM of 20,30,40:
20=2x2x5
30=3x2x5
40=2x2x2x5
So LCM =2x2x5x3x2x5x2x2x2x5
=24000.
The deduct 69 from 24000 to get 23931,which the required number.
Check: 23931/20=1,196.55+7R
23931/30=797.7+7R
23931/40=598.275+7R .
20-7=13
30-7=23
40-7=33
Let us add all these numbers 13+23+33=69
Next find the LCM of 20,30,40:
20=2x2x5
30=3x2x5
40=2x2x2x5
So LCM =2x2x5x3x2x5x2x2x2x5
=24000.
The deduct 69 from 24000 to get 23931,which the required number.
Check: 23931/20=1,196.55+7R
23931/30=797.7+7R
23931/40=598.275+7R .
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