Time Today Is 1PM, What Was The Time 7 Hours Ago?
Example 4-3: If the time today is 1PM, what was the time 7 hours ago?
Solution: First, we need to consider that 1PM has passed 12 o’clock. Therefore, we need to add 1 hour to 12 hours. This means that 1PM is actually 13 hours. Second, we also need to subtract 7 from 13. As such13-7=6. But six has not reached 12 o’clock, thus it is still in the morning. Hence, the time is 6 in the morning or 6AM.
The aforementioned examples illustrate time-of-the-day arithmetic. Similarly, we can also apply the same reasoning that of day-of-the- week arithmetic. If we do this, we need to associate the days of the week as numbers.
The following shows the corresponding numbers of each day of the week Monday – 1 Friday - 5 Tuesday – 2 Saturday - 6 …show more content…
To understand why the value does not change, consider 5≡0 mod 5. In modulo 5 arithmetic, 5 is equivalent to 0. Hence, adding 0 to a number does not change the number. In like manner, adding the modulus of a number does not change the number. Let us consider example 4-7, to illustrate this concept.
Example 4-7: Suppose, we have the congruence15≡6 mod 9. This congruence is equivalent to 15+9=24≡6 mod 9. Note that 15-6 and 24-6 are both divisible by 9, hence the two are congruent.
Problems involving subtraction and multiplication can also be performed. To illustrate these operations, consider example 4-8:
Example 4-8: Evaluate the following: (40-6)mod12 (8-14)mod 4 (8∙5)mod 13
Solution:
Subtract 40-6=34. The result is positive. Divide the difference by the modulus, 12 and take the remainder as the answer. 34÷12 is 2 and the remainder is 10. Hence, the answer is 10. Subtract 8-14=-6. The result is negative. Thus, we need to find x such that (-6-x)/4 is an integer. To do this, try all whole numbers less than the given modulo 4. Among the whole numbers less than 4, we find that when x=2 we have (-6-2)/4=(-8)/4=-2. Hence (8-14) mod 4≡2 Multiply 8∙5=40. Then divide the product by the modulus, 13 and take the remainder as the answer. 40÷13=3 and its remainder is 1. Hence, the desired answer is
Solution: First, we need to consider that 1PM has passed 12 o’clock. Therefore, we need to add 1 hour to 12 hours. This means that 1PM is actually 13 hours. Second, we also need to subtract 7 from 13. As such13-7=6. But six has not reached 12 o’clock, thus it is still in the morning. Hence, the time is 6 in the morning or 6AM.
The aforementioned examples illustrate time-of-the-day arithmetic. Similarly, we can also apply the same reasoning that of day-of-the- week arithmetic. If we do this, we need to associate the days of the week as numbers.
The following shows the corresponding numbers of each day of the week Monday – 1 Friday - 5 Tuesday – 2 Saturday - 6 …show more content…
To understand why the value does not change, consider 5≡0 mod 5. In modulo 5 arithmetic, 5 is equivalent to 0. Hence, adding 0 to a number does not change the number. In like manner, adding the modulus of a number does not change the number. Let us consider example 4-7, to illustrate this concept.
Example 4-7: Suppose, we have the congruence15≡6 mod 9. This congruence is equivalent to 15+9=24≡6 mod 9. Note that 15-6 and 24-6 are both divisible by 9, hence the two are congruent.
Problems involving subtraction and multiplication can also be performed. To illustrate these operations, consider example 4-8:
Example 4-8: Evaluate the following: (40-6)mod12 (8-14)mod 4 (8∙5)mod 13
Solution:
Subtract 40-6=34. The result is positive. Divide the difference by the modulus, 12 and take the remainder as the answer. 34÷12 is 2 and the remainder is 10. Hence, the answer is 10. Subtract 8-14=-6. The result is negative. Thus, we need to find x such that (-6-x)/4 is an integer. To do this, try all whole numbers less than the given modulo 4. Among the whole numbers less than 4, we find that when x=2 we have (-6-2)/4=(-8)/4=-2. Hence (8-14) mod 4≡2 Multiply 8∙5=40. Then divide the product by the modulus, 13 and take the remainder as the answer. 40÷13=3 and its remainder is 1. Hence, the desired answer is