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The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.?
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The digit in the tens place is 8 less than the digit in the thousands ...
Problem: Find a 6-digit number where the digit in the tens place is 8 less than the digit in the thousands place, the digit in the ones place is double the digit in the tens place, the digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places, the digit in the hundreds place is the sum of the digits in the tens and ones places, and all the digits add up to 27.
Solution:
To solve this problem, we can use algebraic equations to represent each of the given conditions. Let's use the variables a, b, c, d, e, and f to represent the digits in the hundred thousands, ten thousands, thousands, hundreds, tens, and ones places, respectively.
1. The digit in the tens place is 8 less than the digit in the thousands place.
b = c - 8
2. The digit in the ones place is double the digit in the tens place.
f = 2b
3. The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places.
a = (b + c)/2
4. The digit in the hundreds place is the sum of the digits in the tens and ones places.
d = b + f
5. All the digits add up to 27.
a + b + c + d + e + f = 27
Now we can use these equations to solve for the digits. We can start with equation 5, which tells us that the sum of all the digits is 27. We can use this to make an equation for one of the variables in terms of the others. For example, we can solve for f in terms of the other variables:
a + b + c + d + e + f = 27
f = 27 - (a + b + c + d + e)
Next, we can substitute this expression for f into equation 2 to get an equation for b in terms of the other variables:
f = 2b
27 - (a + b + c + d + e) = 2b
b = (27 - a - c - d - e)/2
Now we can substitute this expression for b into equations 1, 3, and 4 to get equations for a, c, and d in terms of the other variables:
b = c - 8
(27 - a - c - d - e)/2 = c - 8
d = (27 - a - 2c - e)/2
Finally, we can use these equations to solve for the digits. We can start by picking a value for one of the digits, say e = 3. Then we can use equations 1, 3, and 4 to solve for b, c, and d:
b = (27 - a - c - d - e)/2 = (27 - a - c - d - 3)/2
a = (b + c)/2
d = (27 - a - 2c - e)/2
We can then use equations 1 and 2 to solve for f and e:
f = 2b
e = f/2
Once we have values for all the digits, we can check that they satisfy all the given conditions. For example,
Solution:
To solve this problem, we can use algebraic equations to represent each of the given conditions. Let's use the variables a, b, c, d, e, and f to represent the digits in the hundred thousands, ten thousands, thousands, hundreds, tens, and ones places, respectively.
1. The digit in the tens place is 8 less than the digit in the thousands place.
b = c - 8
2. The digit in the ones place is double the digit in the tens place.
f = 2b
3. The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places.
a = (b + c)/2
4. The digit in the hundreds place is the sum of the digits in the tens and ones places.
d = b + f
5. All the digits add up to 27.
a + b + c + d + e + f = 27
Now we can use these equations to solve for the digits. We can start with equation 5, which tells us that the sum of all the digits is 27. We can use this to make an equation for one of the variables in terms of the others. For example, we can solve for f in terms of the other variables:
a + b + c + d + e + f = 27
f = 27 - (a + b + c + d + e)
Next, we can substitute this expression for f into equation 2 to get an equation for b in terms of the other variables:
f = 2b
27 - (a + b + c + d + e) = 2b
b = (27 - a - c - d - e)/2
Now we can substitute this expression for b into equations 1, 3, and 4 to get equations for a, c, and d in terms of the other variables:
b = c - 8
(27 - a - c - d - e)/2 = c - 8
d = (27 - a - 2c - e)/2
Finally, we can use these equations to solve for the digits. We can start by picking a value for one of the digits, say e = 3. Then we can use equations 1, 3, and 4 to solve for b, c, and d:
b = (27 - a - c - d - e)/2 = (27 - a - c - d - 3)/2
a = (b + c)/2
d = (27 - a - 2c - e)/2
We can then use equations 1 and 2 to solve for f and e:
f = 2b
e = f/2
Once we have values for all the digits, we can check that they satisfy all the given conditions. For example,
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The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.?
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The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.? for Class 5 2024 is part of Class 5 preparation. The Question and answers have been prepared according to the Class 5 exam syllabus. Information about The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.? covers all topics & solutions for Class 5 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.?.
Solutions for The digit in the tens place is 8 less than the digit in the thousands place. The digit in the ones place is double the digit in the tens place. The number is The digit in the hundred thousands place is half of the sum of the digits in the tens and thousands places. The digit in the hundreds place is the sum of the digits in the tens and ones places. All the digits add up to 27.? in English & in Hindi are available as part of our courses for Class 5.
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