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Sagot :
To solve the problem of [tex]\( 5 \longdiv { 7592 } \)[/tex] using long division, follow these steps:
1. Divide 7 by 5:
- 7 divided by 5 is 1 with a remainder of 2.
- Write 1 above the 7. The intermediate remainder is 2.
2. Bring down the next digit (5):
- Now we have 25.
- 25 divided by 5 is 5 with no remainder.
- Write 5 above the 25. The intermediate quotient is now 15.
3. Bring down the next digit (9):
- Now we have 9.
- 9 divided by 5 is 1 with a remainder of 4.
- Write 1 above the 9. The intermediate quotient is now 151.
4. Bring down the last digit (2):
- Now we have 42.
- 42 divided by 5 is 8 with a remainder of 2.
- Write 8 above the 42. The final quotient is now 1518.
5. Remainder:
- Since the remaining number is 2 and there are no more digits to bring down, 2 is our remainder.
Thus, the quotient of [tex]\( 7592 \)[/tex] divided by [tex]\( 5 \)[/tex] is [tex]\( 1518 \)[/tex] with a remainder of [tex]\( 2 \)[/tex]. So the final answer is:
[tex]\[ \boxed{1518 \text{ R } 2} \][/tex]
1. Divide 7 by 5:
- 7 divided by 5 is 1 with a remainder of 2.
- Write 1 above the 7. The intermediate remainder is 2.
2. Bring down the next digit (5):
- Now we have 25.
- 25 divided by 5 is 5 with no remainder.
- Write 5 above the 25. The intermediate quotient is now 15.
3. Bring down the next digit (9):
- Now we have 9.
- 9 divided by 5 is 1 with a remainder of 4.
- Write 1 above the 9. The intermediate quotient is now 151.
4. Bring down the last digit (2):
- Now we have 42.
- 42 divided by 5 is 8 with a remainder of 2.
- Write 8 above the 42. The final quotient is now 1518.
5. Remainder:
- Since the remaining number is 2 and there are no more digits to bring down, 2 is our remainder.
Thus, the quotient of [tex]\( 7592 \)[/tex] divided by [tex]\( 5 \)[/tex] is [tex]\( 1518 \)[/tex] with a remainder of [tex]\( 2 \)[/tex]. So the final answer is:
[tex]\[ \boxed{1518 \text{ R } 2} \][/tex]
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