Can you solve it? Carl Friedrich Gauss, money saving expert (2024)

Financially well-informed readers may have recently learned about a money-saving trick that uses a formula supposedly devised by Carl Friedrich Gauss, the 19th century maths colossus. The viral 100 envelope challenge is based on an apocryphal story that Gauss, when a young boy, outwitted his teacher by adding the numbers from 1 to 100 almost instantly.

The child genius had realised that if you group the numbers from 1 to 100 in pairs, the sum is equal to (1 + 100) + (2 + 99) + (3 + 98) + … In other words, 101 + 101 + 101 + … Since there are 50 pairs of numbers, the sum is 101 x 50 = 5050.

The 100 envelope challenge is a way of saving £5050 in 25 weeks. You mark up one hundred envelopes with the amounts from £1 to £100. Twice a week you choose two envelopes at random, fill each envelope with the amount it says, and put them aside. After 25 weeks, all the envelopes will be full. Bang! More than 5K saved.

Of course, the challenge presupposes that you have £5050 to save in the first place and that you possess a large number of coins and banknotes. Still, the tens of thousands of reactions on social media to the trick seem to indicate that many people find it inspiring. “If you have a three-year plan to buy a house, you could have a little over £30,000 for a down payment by doing this,” one saver reportedly posted on social media. Perhaps the inclusion of randomness is what makes the exercise more fun.

Today’s puzzles require you to summon your inner Gauss and find fast ways of counting number lists. You are not allowed to use a calculator nor add them all up individually!

1. The double bill

Which of the two additions below (the sum of the nine numbers on the left, and the sum of the nine on the right) adds to the highest number?

Can you solve it? Carl Friedrich Gauss, money saving expert (1)

2. The 2D envelope challenge

Sum all the numbers in the grid below.

Can you solve it? Carl Friedrich Gauss, money saving expert (2)

Please NO SPOILERS

I’ll be back at 5pm UK with the solutions.

UPDATE: To read the solutions click here.

While you are here, can I go off topic briefly and ask you to spread the word about the Young Sportswriter of the Year award, run by the Guardian and Football School, the kids book series I co-write with Ben Lyttleton. Children between 7-12 need to submit up to 600 words (before August) about a sportsperson or sports event and can win fantastic prizes, such as their story published by the Guardian.

Can you solve it? Carl Friedrich Gauss, money saving expert (3)

A fantastic line up of judges includes Tour de France cyclist Tao Geoghegan Hart, Olympic gold medallist Christine Ohuruogu, golfer Ian Poulter and German World Cup winner Per Mertesacker. For more information please click here. With the Euros, Wimbledon, the Tour de France, and imminently the Olympics, there is no shortage of subject matter!

Can you solve it? Carl Friedrich Gauss, money saving expert (4)

I’m the author of several books of maths and puzzles and also of the book series Football School. The latest book in the series, Terrific Teams, is out this week. It contains stories about 50 amazing teams from Bungay Town to Brazil, and from Lewes to Liverpool.

Sources for today’s puzzles: 1) The Moscow Puzzles by Boris A Kordemsky, 2) Algorithmic Puzzles by Anany and Maria Levitin.

Can you solve it? Carl Friedrich Gauss, money saving expert (2024)

FAQs

What is the trick of the Gauss formula? ›

Young Gauss's “trick” for finding the sum of an arithmetic progression is usually explained in terms of adding pairs of elements from opposite ends of the sequence, so that all the pairs have the same sum. One way to envision this process is to fold the series in half with a hairpin bend.

Who was Carl Friedrich Gauss short paragraph? ›

Johann Friedrich Carl Gauss was born in 1777 to a poor family in Brunswick, Germany. The boy was found to be a mathematical prodigy. Gauss' amazing calculating abilities aroused the interest of his teachers, and the child received a solid education despite lack of money.

What is the 100 envelope challenge equation? ›

Each envelope will have a certain amount of cash, from 1 to 100. When you add all the envelopes together — $100 + $99 + 98 + $97, and so on, all the way down to $1 — the total amount comes out to $5,050. If you were to save the same amount each day for 100 days, you'd need to set aside $50.50 per day to hit that sum.

What is the sum of the numbers from 1 to 100 Gauss? ›

Gauss used this same method to sum all the numbers from 1 to 100. He realized that he could pair up all the numbers. That meant he had 50 pairs, each with a sum of 101. He could then multiply 50 x 101 to arrive at his answer: 5050.

What was Gauss' IQ? ›

Carl Gauss was a child prodigy who went on to become one of the greatest German mathematicians of the 19th century. With IQ scores ranging from 250 to 300, he contributed to the number theory fields like algebra, statistics, and analysis.

What was Carl Gauss conclusion in math? ›

In his doctoral thesis from 1799, Gauss proved the fundamental theorem of algebra which states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

What did Carl Friedrich Gauss prove? ›

His doctoral thesis of 1797 gave a proof of the fundamental theorem of algebra: every polynomial equation with real or complex coefficients has as many roots (solutions) as its degree (the highest power of the variable). Gauss's proof, though not wholly convincing, was remarkable for its critique of earlier attempts.

How to save $5000 in 3 months? ›

How To Save $5,000 in 3 Months
  1. Create a budget.
  2. Find ways to increase your income.
  3. Reduce expenses.
  4. Embrace savings challenges.
  5. Automate your savings.
  6. Track your progress.
Aug 17, 2024

How to save $5000 in 100 days? ›

The 100 Envelope Saving Challenge provides a structured system to save over $5,000 in 100 days. To begin, gather 100 envelopes and sequentially number them $1 through $100. On day one, you'll place $1 in envelope one. On day two, you'll add $2 to envelope two, and so on.

How to save 10k in 3 months envelope? ›

On each envelope, write the day number and the amount you need to save for that day. For instance, on the first envelope, you would write "Day 1: $1" and on the second envelope "Day 2: $2", and so on all the way to Day 100: $100. Each day, you take the envelope for that day and put the designated amount of cash inside.

What is 1 * 2 * 3 * 4 * 5 all the way to 100? ›

The answer is 5050, it's a programming exercise I sometimes set. Mathematically though you can do this in your head far faster than you can write a program to do it. How many of these sums are there? Clearly there are 100/2 = 50 sums.

What is Gauss's formula? ›

The electric field is perpendicular, locally, to the equipotential surface of the conductor, and zero inside; its flux πa2·E, by Gauss's law equals πa2·σ/ε0. Thus, σ = ε0E.

What is Gauss famous for? ›

Born 241 years ago on April 30, Johann Carl Friedrich Gauss is often described as the “Prince of Mathematicians” and hailed for his contributions to number theory, geometry, probability theory and astronomy.

How to do Gauss formula? ›

Simply list the ENTIRE series forward, then list the entire series in reverse and add the pairs. In this situation, you will need to multiply the sum by the number of pairs and then divide by two, since you are actually working with 2 complete series.

What is the Gauss theorem for dummies? ›

Gauss's law for electricity states that the electric flux Φ across any closed surface is proportional to the net electric charge q enclosed by the surface; that is, Φ = q/ε0, where ε0 is the electric permittivity of free space and has a value of 8.854 × 1012 square coulombs per newton per square metre.

How do you calculate Gauss? ›

The flux Φ of the electric field →E through any closed surface S (a Gaussian surface) is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ϵ0): Φ=∮S→E⋅ˆndA=qencϵ0.

How do you solve Gauss theorem? ›

Where, 2Πrl is the curved surface area of the cylinder.
  1. Therefore, ϕE=qϵ0.
  2. Charge enclosed in cylinder = line charge density × length= λl. E(2πrl)=lλϵ0.
  3. Or, E=λ(2πϵ0r)
  4. Using Gauss theorem, ϕ=∮→E→dA=qϵo.
  5. Φ = EA + EA = 2EA. E=σA2ϵo.
Oct 15, 2023

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